Practice Full Course (ID 17)

Let $h(t) = t^2 e^{-t^6}$. Let's check if it's even or odd.
$h(-t) = (-t)^2 e^{-(-t)^6} = t^2 e^{-t^6} = h(t)$
Since $h(-t) = h(t)$, the integrand $h(t)$ is an even function.




Consider $f(x) = \int_{-x}^{0} h(t) dt$. Let's use the substitution $u = -t$.
This means $t = -u$ and $dt = -du$.
We also change the limits of integration:
When $t = -x$, $u = -(-x) = x$.
When $t = 0$, $u = -0 = 0$.

Substitute into the integral for $f(x)$:
$f(x) = \int_{u=x}^{u=0} h(-u) (-du)$
Since $h(t)$ is even, $h(-u) = h(u)$.
$f(x) = \int_{x}^{0} h(u) (-du)$
Using the property $\int_a^b = -\int_b^a$, we flip the limits and remove the negative sign from $-du$:
$f(x) = \int_{0}^{x} h(u) du$

Now compare this to $g(x) = \int_{0}^{x} h(t) dt$. Changing the variable of integration from $t$ to $u$ doesn't change the value.
Therefore, $f(x) = \int_{0}^{x} h(u) du = g(x)$.

So, $f(x) = g(x)$ for all $x$ in the domain.




Based on the finding that $f(x) = g(x)$:

(a) $f(x) = g(x)$ for all $x \in [0, \infty)$. This is TRUE.

(b) $f'(x) = -g'(x)$ for all $x \in [0, \infty)$. Since $f(x)=g(x)$, their derivatives must be equal: $f'(x)=g'(x)$. Let's find $g'(x)$ using FTC Part 1: $g'(x) = \frac{d}{dx} \int_{0}^{x} t^2 e^{-t^6} dt = x^2 e^{-x^6}$. So $f'(x) = x^2 e^{-x^6}$. The statement $f'(x)=-g'(x)$ would imply $x^2 e^{-x^6} = -x^2 e^{-x^6}$, which is only true if $x=0$. Thus, the statement is FALSE for $x>0$.

(c) $f(x)$ is decreasing and $g(x)$ is increasing on $[0, \infty)$. We found $f'(x) = g'(x) = x^2 e^{-x^6}$. Since $x^2 \ge 0$ and $e^{-x^6} > 0$, both derivatives are non-negative ($\ge 0$). Therefore, both functions are increasing (or non-decreasing). FALSE.

(d) $f(x) = 0$ for all $x \in [0, \infty)$. This implies $g(x)=0$. But $g(x) = \int_{0}^{x} t^2 e^{-t^6} dt$. The integrand $t^2 e^{-t^6}$ is strictly positive for $t \ne 0$. For $x > 0$, $g(x)$ is the integral of a positive function over an interval of positive length, so $g(x) > 0$. FALSE.

(e) $f(x) < 0$ for all $x \in (0, \infty)$. Since $f(x)=g(x)$ and we found $g(x) > 0$ for $x > 0$, this is FALSE.

The relationship between integrals of symmetric functions over symmetric limits is key. For an EVEN function $h(t)$, $\int_{-x}^0 h(t) dt = \int_0^x h(t) dt$. For an ODD function $h(t)$, $\int_{-x}^0 h(t) dt = -\int_0^x h(t) dt$.

The only true statement is (a).

$$\boxed{\text{Statement (a) is true: } f(x) = g(x)}$$

A function $f$ is odd if $f(-x) = -f(x)$ for all $x$.
A key property for definite integrals of odd functions is:
For any odd continuous function $f$, $\int_{-c}^{c} f(x) dx = 0$ for any $c$ in its domain.
We are asked to evaluate $\int_{-a}^{b} f(x) dx$ where $0 < a < b$.




We can split the interval of integration $[-a, b]$ into two parts in a way that lets us use the property mentioned above. Since $-a < a < b$, we can split the integral at $x = a$:
$\int_{-a}^{b} f(x) dx = \int_{-a}^{a} f(x) dx + \int_{a}^{b} f(x) dx$




The first part of the split integral is $\int_{-a}^{a} f(x) dx$.
Since $f(x)$ is an odd function, we know that:
$\int_{-a}^{a} f(x) dx = 0$




Substitute the result from Step 3 back into the split integral equation from Step 2:
$\int_{-a}^{b} f(x) dx = 0 + \int_{a}^{b} f(x) dx$
$\int_{-a}^{b} f(x) dx = \int_{a}^{b} f(x) dx$

This result matches option (c).



$$\begin{array}{rl}{\displaystyle {\int }_{-a}^{b}f(x)dx}& {\displaystyle ={\int }_{-a}^{a}f(x)dx+{\int }_a^{b}f(x)dx}\\ {\displaystyle }& {\displaystyle =0+{\int }_a^{b}f(x)dx(\text{since }f\text{ is odd})}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle {\int }_a^{b}f(x)dx}}}\end{array}$$
\begin{align*}
\int_{-a}^{b} f(x) dx &= \int_{-a}^{a} f(x) dx + \int_{a}^{b} f(x) dx \\
&= 0 + \int_{a}^{b} f(x) dx \quad (\text{since } f \text{ is odd}) \\
&= \boxed{\int_{a}^{b} f(x) dx}
\end{align*}

We need the area of the region bounded by the curve $y = \sqrt[3]{x}$, the y-axis (which is the line $x = 0$), and the horizontal line $y = 3$.
The curve $y = \sqrt[3]{x}$ passes through the origin $(0, 0)$.
The region is bounded on the left by $x = 0$, on the right by the curve, below by $y = 0$ (where the curve meets the y-axis), and above by $y = 3$.

Since the region is defined by boundaries involving $y$, it's convenient to integrate with respect to y.




We need to express the right boundary curve in the form $x = f(y)$.

Given $y = \sqrt[3]{x}$, we cube both sides to solve for $x$:
$x = y^3$

The left boundary is the y-axis, which has the equation $x = 0$.




When integrating with respect to y, the area between a right curve $x = f(y)$ and a left curve $x = g(y)$ from $y = c$ to $y = d$ is given by:
$A = \int_c^d [f(y) - g(y)] dy$

In our case:
The right curve is $x_{right} = y^3$.
The left curve is $x_{left} = 0$.
The limits of integration are the y-values defining the region's vertical extent, which are from $y = 0$ to $y = 3$.

So, the integral for the area $A$ is:
$A = \int_0^3 (y^3 - 0) dy$




We evaluate the integral $\int_0^3 y^3 dy$.
The antiderivative of $y^3$ is $\frac{y^4}{4}$.
We evaluate this from $y=0$ to $y=3$.

The result $81/4$ corresponds to option (d).



$$\begin{array}{rl}{\displaystyle A}& {\displaystyle ={\int }_0^3(y^3-0)dy}\\ {\displaystyle }& {\displaystyle ={\int }_0^3{y}^{3}dy}\\ {\displaystyle }& {\displaystyle ={\left[\frac{y^4}{4}\right]}_0^3}\\ {\displaystyle }& {\displaystyle =\left(\frac{3^4}{4}\right)-\left(\frac{0^4}{4}\right)}\\ {\displaystyle }& {\displaystyle =\frac{81}{4}-0}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle \frac{81}{4}}}}\end{array}$$
\begin{align*}
A &= \int_0^3 (y^3 - 0) dy \\
&= \int_0^3 y^3 dy \\
&= \left[ \frac{y^4}{4} \right]_0^3 \\
&= \left( \frac{3^4}{4} \right) - \left( \frac{0^4}{4} \right) \\
&= \frac{81}{4} - 0 \\
&= \boxed{\frac{81}{4}}
\end{align*}

The expression $(r^2 - x^2)$ appears inside the integral. Recall the equation of a circle centered at the origin with radius $r$: $x^2 + y^2 = r^2$. The upper semi-circle is given by $y = \sqrt{r^2 - x^2}$.

The integral involves $\pi (r^2 - x^2)$, which looks related to the disk method for finding volumes of revolution.

The volume of a solid generated by revolving the area under a curve $y = f(x)$ from $x=a$ to $x=b$ around the x-axis is given by the disk method formula: $V = \int_a^b \pi [f(x)]^2 dx$.

Let's consider the function $y = \sqrt{r^2 - x^2}$. If we revolve the area under this curve from $x=0$ to $x=r$ (which is a quarter-circle) around the x-axis, the volume generated is a hemisphere. Using the disk method, its volume would be:
$V_{hemi} = \int_0^r \pi (\sqrt{r^2 - x^2})^2 dx = \int_0^r \pi (r^2 - x^2) dx$




The integral in the question is $\int_0^r 2\pi (r^2 - x^2) dx$.
We can factor out the 2: $2 \int_0^r \pi (r^2 - x^2) dx$.

Notice that $\int_0^r \pi (r^2 - x^2) dx$ is exactly the integral we identified as the volume of a hemisphere of radius $r$.

Therefore, the given integral is $2 \times V_{hemi}$.
Two hemispheres make a full sphere. So, the integral represents the volume of a sphere of radius $r$.




We can directly calculate the integral to verify. The volume of a sphere is known to be $\frac{4}{3}\pi r^3$. Let's see if the integral evaluates to this.

The calculation shows that the integral indeed equals the formula for the volume of a sphere. Therefore, the correct option is (a).



$$\begin{array}{rl}{\displaystyle I}& {\displaystyle ={\int }_0^{r}2\pi (r^2-x^2)dx}\\ {\displaystyle }& {\displaystyle =2\pi {\int }_0^r(r^2-x^2)dx}\\ {\displaystyle }& {\displaystyle =2\pi {[r^{2}x-\frac{x^3}{3}]}_0^r}\\ {\displaystyle }& {\displaystyle =2\pi ((r^2(r)-\frac{r^3}{3})-(r^2(0)-\frac{0^3}{3}))}\\ {\displaystyle }& {\displaystyle =2\pi (r^3-\frac{r^3}{3}-0)}\\ {\displaystyle }& {\displaystyle =2\pi \left(\frac{3r^3-r^3}{3}\right)}\\ {\displaystyle }& {\displaystyle =2\pi \left(\frac{2r^3}{3}\right)}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle \frac{4}{3}\pi r^3}}}\end{array}$$
\begin{align*}
I &= \int_0^r 2\pi (r^2 - x^2) dx \\
&= 2\pi \int_0^r (r^2 - x^2) dx \\
&= 2\pi \left[ r^2x - \frac{x^3}{3} \right]_0^r \\
&= 2\pi \left( (r^2(r) - \frac{r^3}{3}) - (r^2(0) - \frac{0^3}{3}) \right) \\
&= 2\pi \left( r^3 - \frac{r^3}{3} - 0 \right) \\
&= 2\pi \left( \frac{3r^3 - r^3}{3} \right) \\
&= 2\pi \left( \frac{2r^3}{3} \right) \\
&= \boxed{\frac{4}{3}\pi r^3}
\end{align*}

The average value of a function $f(x)$ over an interval $[a, b]$ is given by the integral of the function over that interval, divided by the length of the interval.

The formula for the average value $f_{avg}$ of a function $f$ on the interval $[a, b]$ is:
$f_{avg} = \frac{1}{b-a} \int_a^b f(x) dx$

For our specific function $g(x) = \sin(x)$ and the interval $[0, \pi]$, we have $a=0$ and $b=\pi$.

So, we need to calculate:
$g_{avg} = \frac{1}{\pi - 0} \int_0^{\pi} \sin(x) dx$




First, we find the antiderivative of $\sin(x)$. The integral of $\sin(x)$ is $-\cos(x)$.

Next, we evaluate the definite integral $\int_0^{\pi} \sin(x) dx$ using the Fundamental Theorem of Calculus.

We evaluate the antiderivative $-\cos(x)$ at the upper limit ($x=\pi$) and subtract its value at the lower limit ($x=0$).




Now we substitute the value of the integral back into the average value formula.

Remember that the length of the interval is $b-a = \pi - 0 = \pi$.

Divide the result of the integral (which is 2) by the length of the interval ($\pi$) to get the average value.


$$\begin{array}{rl}{\displaystyle g_{avg}}& {\displaystyle =\frac{1}{\pi -0}{\int }_0^{\pi }\sin (x)dx}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }{\int }_0^{\pi }\sin (x)dx}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }{[-\cos (x)]}_0^{\pi }}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }((-\cos (\pi ))-(-\cos (0)))}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }((-(-1))-(-(1)))}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }(1-(-1))}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }(1+1)}\\ {\displaystyle }& {\displaystyle =\frac{1}{\pi }(2)}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle \frac{2}{\pi }}}}\end{array}$$
\begin{align*}
g_{avg} &= \frac{1}{\pi - 0} \int_0^{\pi} \sin(x) dx \\
&= \frac{1}{\pi} \int_0^{\pi} \sin(x) dx \\
&= \frac{1}{\pi} \left[ -\cos(x) \right]_0^{\pi} \\
&= \frac{1}{\pi} \left( (-\cos(\pi)) - (-\cos(0)) \right) \\
&= \frac{1}{\pi} \left( (-(-1)) - (-(1)) \right) \\
&= \frac{1}{\pi} \left( 1 - (-1) \right) \\
&= \frac{1}{\pi} (1 + 1) \\
&= \frac{1}{\pi} (2) \\
&= \boxed{\frac{2}{\pi}}
\end{align*}

Using the product rule on the left side:

$$\frac{d}{dx}((1+x+x^2)e^p)=\frac{d}{dx}(1)$$
\frac{d}{dx} \left( (1 + x + x^2) e^p \right) = \frac{d}{dx}(1)


The derivative of the right side is:

$$\frac{d}{dx}(1)=0$$
\frac{d}{dx}(1) = 0


Now apply the product rule to the left side:

$$\frac{d}{dx}(1+x+x^2)\cdot e^p+(1+x+x^2)\cdot \frac{d}{dx}(e^p)=0$$
\frac{d}{dx} \left( 1 + x + x^2 \right) \cdot e^p + (1 + x + x^2) \cdot \frac{d}{dx}(e^p) = 0


This simplifies to:

$$(1+2x)e^p+(1+x+x^2)e^p\frac{dp}{dx}=0$$
(1 + 2x) e^p + (1 + x + x^2) e^p \frac{dp}{dx} = 0




First, isolate the term with $\frac{dp}{dx}$:

$$(1+x+x^2)e^p\frac{dp}{dx}=-(1+2x)e^p$$
(1 + x + x^2) e^p \frac{dp}{dx} = -(1 + 2x) e^p


Now, divide both sides by $(1 + x + x^2) e^p$ to get:

$$\frac{dp}{dx}=-\frac{1+2x}{1+x+x^2}$$
\frac{dp}{dx} = -\frac{1 + 2x}{1 + x + x^2}




When $x = 1$, the expression becomes:

$$\frac{dp}{dx}=-\frac{1+2(1)}{1+1+1^2}=-\frac{1+2}{1+1+1}=-\frac{3}{3}=-1$$
\frac{dp}{dx} = -\frac{1 + 2(1)}{1 + 1 + 1^2} = -\frac{1 + 2}{1 + 1 + 1} = -\frac{3}{3} = -1




Thus, the rate of change $\frac{dp}{dx}$ when $x = 1$ is:

$$\boxed{{\displaystyle -1}}$$
\boxed{-1}


The correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}-1}}$$
\boxed{(A) \, -1}

To evaluate $g'(3)$, we first need to find the derivative of $g(x) = \sqrt{f(x)}$. We can rewrite this as $g(x) = (f(x))^{1/2}$. We must use the chain rule to find the derivative.
$$\begin{array}{rl}{\displaystyle }& {\displaystyle g^{\mathrm{\prime }}(x)}\\ {\displaystyle }& {\displaystyle =\frac{1}{2}(f(x){)}^{-1\mathrm{/}2}\cdot f^{\mathrm{\prime }}(x)}\\ {\displaystyle }& {\displaystyle =\frac{f^{\mathrm{\prime }}(x)}{2\sqrt{f(x)}}}\end{array}$$
\begin{aligned}
& \quad g'(x) \\
&= \frac{1}{2}(f(x))^{-1/2} \cdot f'(x) \\
&= \frac{f'(x)}{2\sqrt{f(x)}}
\end{aligned}




Next, we need to find two values from the provided graph: the value of the function $f(3)$ and the slope of the tangent line $f'(3)$ at $x=3$.

From the graph, we can see that the red curve passes through the point $(3, 2)$.
$$\begin{array}{rl}{\displaystyle }& {\displaystyle f(3)}\\ {\displaystyle }& {\displaystyle =2}\end{array}$$
\begin{aligned}
& \quad f(3) \\
&= 2
\end{aligned}

The blue line is the tangent line to the curve at $x=3$. We can find its slope by identifying two points on the line, for example, $(0, 4)$ and $(3, 2)$.
$$\begin{array}{rl}{\displaystyle }& {\displaystyle f^{\mathrm{\prime }}(3)}\\ {\displaystyle }& {\displaystyle =\frac{2-4}{3-0}}\\ {\displaystyle }& {\displaystyle =\frac{-2}{3}}\end{array}$$
\begin{aligned}
& \quad f'(3) \\
&= \frac{2 - 4}{3 - 0} \\
&= \frac{-2}{3}
\end{aligned}




Now we substitute the values we found into the derivative formula from Step 1.
$$\begin{array}{rl}{\displaystyle }& {\displaystyle g^{\mathrm{\prime }}(3)}\\ {\displaystyle }& {\displaystyle =\frac{f^{\mathrm{\prime }}(3)}{2\sqrt{f(3)}}}\\ {\displaystyle }& {\displaystyle =\frac{-2\mathrm{/}3}{2\sqrt{2}}}\\ {\displaystyle }& {\displaystyle =\frac{-2\mathrm{/}3}{2\sqrt{2}}}\\ {\displaystyle }& {\displaystyle =\frac{-1}{3\sqrt{2}}}\\ {\displaystyle }& {\displaystyle =\frac{-1}{3\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}}\\ {\displaystyle }& {\displaystyle =-\frac{\sqrt{2}}{6}}\end{array}$$
\begin{aligned}
& \quad g'(3) \\
&= \frac{f'(3)}{2\sqrt{f(3)}} \\
&= \frac{-2/3}{2\sqrt{2}} \\
&= \frac{-2/3}{2\sqrt{2}} \\
&= \frac{-1}{3\sqrt{2}} \\
&= \frac{-1}{3\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \\
&= -\frac{\sqrt{2}}{6}
\end{aligned}




$$\begin{array}{rl}{\displaystyle }& {\displaystyle =\boxed{{\displaystyle -\frac{\sqrt{2}}{6}}}}\end{array}$$
\begin{aligned}
&= \boxed{-\frac{\sqrt{2}}{6}}
\end{aligned}

First, notice that as $t \to \infty$, the expression inside the parentheses approaches 1, while the exponent approaches $\infty$. This gives the indeterminate form $1^\infty$, which often suggests that the limit involves the exponential function, $e$. Our goal is to manipulate the expression to fit the standard form for the definition of $e$. We can do this by rewriting the base, $\frac{t+4}{t-1}$, into the form $1 + \text{something}$.


By rewriting $t+4$ as $t-1+5$, we can split the fraction. This isolates a '1' which is crucial for the next step.

$$\begin{align*}$$
& \lim_{t \to \infty} \left(\frac{t+4}{t-1}\right)^{t} \\
&= \lim_{t \to \infty} \left(\frac{t-1+5}{t-1}\right)^{t} \\
&= \lim_{t \to \infty} \left(\frac{t-1}{t-1} + \frac{5}{t-1}\right)^{t} \\
&= \lim_{t \to \infty} \left(1 + \frac{5}{t-1}\right)^{t}
\end{align*}

The key formula to remember is the limit definition of the exponential function: $\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x$. Our entire strategy is to make our limit look like this form.
Now that our limit looks very similar to the definition, we can use a substitution to make it match perfectly.


Let $n = t-1$. As $t \to \infty$, it's also true that $n \to \infty$. From this substitution, we also have $t = n+1$. After substituting, we can split the exponent using the property $a^{b+c} = a^b \cdot a^c$. This allows us to separate the expression into two parts. The first part will match the definition of $e^x$ exactly, and the second part will be a simple limit to evaluate.

$$\begin{align*}$$
& \text{Let } n = t-1. \text{ As } t \to \infty, n \to \infty. \\
& \text{Also, } t = n+1. \\
& \lim_{n \to \infty} \left(1 + \frac{5}{n}\right)^{n+1} \\
&= \lim_{n \to \infty} \left[ \left(1 + \frac{5}{n}\right)^{n} \cdot \left(1 + \frac{5}{n}\right)^{1} \right] \\
&= \left[ \lim_{n \to \infty} \left(1 + \frac{5}{n}\right)^{n} \right] \cdot \left[ \lim_{n \to \infty} \left(1 + \frac{5}{n}\right) \right] \\
&= (e^5) \cdot (1+0) \\
&= \boxed{e^5}
\end{align*}

The integrand is $f(x) = \frac{1}{x-1}$.
This function has an infinite discontinuity (a vertical asymptote) when the denominator is zero, i.e., when $x-1 = 0$, which means $x = 1$.
Since this discontinuity occurs at the lower limit of integration, this is a Type 2 improper integral.



To evaluate this improper integral, we replace the problematic lower limit with a variable (say, $c$) and take the limit as $c$ approaches $1$ from the right side (since our interval is $[1, 2]$, we approach 1 from within the interval):

$\int_1^2 \frac{1}{x-1} dx = \lim_{c\to 1^+} \int_c^2 \frac{1}{x-1} dx$



First, we find the antiderivative of $\frac{1}{x-1}$.
Let $u = x-1$, then $du = dx$.
So, $\int \frac{1}{x-1} dx = \int \frac{1}{u} du = \ln|u| = \ln|x-1|$.

Now, evaluate the definite integral from $c$ to $2$:
$\int_c^2 \frac{1}{x-1} dx = \left[ \ln|x-1| \right]_c^2$
$= \ln|2-1| - \ln|c-1|$
$= \ln(1) - \ln|c-1|$
Since $\ln(1) = 0$, this simplifies to:
$= -\ln|c-1|$

Since $c \to 1^+$, $c$ is slightly greater than $1$, so $c-1$ is a small positive number. Thus, $|c-1| = c-1$.
So, the definite integral is $-\ln(c-1)$.



Now we substitute this back into our limit expression:
$L = \lim_{c\to 1^+} \left( -\ln(c-1) \right)$

Let $y = c-1$. As $c \to 1^+$, $y \to 0^+$.
The limit becomes:
$L = \lim_{y\to 0^+} (-\ln(y))$

We know that $\lim_{y\to 0^+} \ln(y) = -\infty$.
Therefore:
$L = -(-\infty) = +\infty$

When evaluating improper integrals with discontinuities, always rewrite them as a limit first. If the limit is $\infty$, $-\infty$, or does not exist, the integral diverges. If the limit is a finite number, the integral converges to that number.

Since the limit is $\infty$, the integral diverges.



$$\begin{array}{rl}{\displaystyle {\int }_1^2\frac{1}{x-1}dx}& {\displaystyle =\underset{c\to 1^{+}}{\lim }{\int }_c^2\frac{1}{x-1}dx}\\ {\displaystyle }\\ {\displaystyle \text{First, find the antiderivative:}}\\ {\displaystyle \int \frac{1}{x-1}dx}& {\displaystyle =\ln \mathrm{\mid }x-1\mathrm{\mid }}\\ {\displaystyle }\\ {\displaystyle \text{Evaluate the definite integral part:}}\\ {\displaystyle {\int }_c^2\frac{1}{x-1}dx}& {\displaystyle ={[\ln \mathrm{\mid }x-1\mathrm{\mid }]}_c^2}\\ {\displaystyle }& {\displaystyle =\ln \mathrm{\mid }2-1\mathrm{\mid }-\ln \mathrm{\mid }c-1\mathrm{\mid }}\\ {\displaystyle }& {\displaystyle =\ln (1)-\ln \mathrm{\mid }c-1\mathrm{\mid }}\\ {\displaystyle }& {\displaystyle =0-\ln \mathrm{\mid }c-1\mathrm{\mid }}\\ {\displaystyle }& {\displaystyle =-\ln \mathrm{\mid }c-1\mathrm{\mid }}\\ {\displaystyle }\\ {\displaystyle \text{Since }c\to 1^{+},c-1>0,\text{ so }\mathrm{\mid }c-1\mathrm{\mid }=c-1:}\\ {\displaystyle }& {\displaystyle =-\ln (c-1)}\\ {\displaystyle }\\ {\displaystyle \text{Now, evaluate the limit:}}\\ {\displaystyle \underset{c\to 1^{+}}{\lim }(-\ln (c-1))}\\ {\displaystyle \text{Let }y=c-1.\text{ As }c\to 1^{+},y\to 0^{+}\mathrm{.}}\\ {\displaystyle \underset{y\to 0^{+}}{\lim }(-\ln (y))}& {\displaystyle =-(-\mathrm{\infty })}\\ {\displaystyle }& {\displaystyle =+\mathrm{\infty }}\\ {\displaystyle }\\ {\displaystyle \text{Therefore, the integral }}& {\displaystyle \boxed{{\displaystyle \text{diverges}}}\mathrm{.}}\end{array}$$
\begin{align*}
\int_1^2 \frac{1}{x-1} dx &= \lim_{c\to 1^+} \int_c^2 \frac{1}{x-1} dx \\
\\
\text{First, find the antiderivative:} \\
\int \frac{1}{x-1} dx &= \ln|x-1| \\
\\
\text{Evaluate the definite integral part:} \\
\int_c^2 \frac{1}{x-1} dx &= \left[ \ln|x-1| \right]_c^2 \\
&= \ln|2-1| - \ln|c-1| \\
&= \ln(1) - \ln|c-1| \\
&= 0 - \ln|c-1| \\
&= -\ln|c-1| \\
\\
\text{Since } c \to 1^+, c-1 > 0, \text{ so } |c-1| = c-1: \\
&= -\ln(c-1) \\
\\
\text{Now, evaluate the limit:} \\
\lim_{c\to 1^+} \left( -\ln(c-1) \right) \\
\text{Let } y = c-1. \text{ As } c \to 1^+, y \to 0^+. \\
\lim_{y\to 0^+} (-\ln(y)) &= -(-\infty) \\
&= +\infty \\
\\
\text{Therefore, the integral } &\boxed{\text{diverges}}.
\end{align*}

The expression $\sqrt{e^x - 1}$ in the denominator is a good candidate for substitution. Let's try substituting for the entire square root to simplify it.

Let $u = \sqrt{e^x - 1}$.
Squaring both sides gives $u^2 = e^x - 1$.
From this, we can express $e^x$ in terms of $u$: $e^x = u^2 + 1$.

Now, we need to find $dx$ in terms of $du$. Differentiate $e^x = u^2 + 1$ implicitly with respect to $x$ for the left side and $u$ for the right side (or solve for $x$ first: $x = \ln(u^2+1)$, then $dx = \frac{2u}{u^2+1} du$).
Using implicit differentiation on $e^x = u^2 + 1$:
$e^x dx = 2u du$.

We also need to express the numerator $e^{2x}$ in terms of $u$.
$e^{2x} = (e^x)^2 = (u^2 + 1)^2$.

Alternatively, we can write the integral as $\int \frac{e^x \cdot e^x dx}{\sqrt{e^x - 1}}$. We already have $e^x dx = 2u du$ and $e^x = u^2 + 1$. This approach seems more direct.

A common strategy for integrals involving $\sqrt{a e^x + b}$ or similar expressions is to let $u$ be the entire square root, or the expression inside the square root. Letting $u = \sqrt{e^x - 1}$ simplifies the denominator directly to $u$.




The integral is $I = \int \frac{e^{2x}}{\sqrt{e^x - 1}} dx = \int \frac{e^x \cdot e^x dx}{\sqrt{e^x - 1}}$.
Substitute the expressions found in Step 1:
Numerator term $e^x = u^2 + 1$.
Denominator $\sqrt{e^x - 1} = u$.
Differential term $e^x dx = 2u du$.

So, the integral becomes:
$I = \int \frac{(u^2 + 1) (2u du)}{u}$




The $u$ in the numerator and denominator cancels out (assuming $u \ne 0$, which is true if $e^x - 1 > 0$, i.e., $x > 0$):
$I = \int 2(u^2 + 1) du$
$I = \int (2u^2 + 2) du$

Now, find the antiderivative using the power rule for integration:
$I = 2 \frac{u^3}{3} + 2u + C$, where $C$ is the constant of integration.
$I = \frac{2}{3}u^3 + 2u + C$




Recall our substitution: $u = \sqrt{e^x - 1} = (e^x - 1)^{1/2}$.
Substitute this back into the expression for $I$:
$I = \frac{2}{3}(\sqrt{e^x - 1})^3 + 2\sqrt{e^x - 1} + C$
$I = \frac{2}{3}(e^x - 1)^{3/2} + 2(e^x - 1)^{1/2} + C$

This is a correct answer, but it can be simplified by factoring out $2(e^x - 1)^{1/2}$:
$I = 2(e^x - 1)^{1/2} \left[ \frac{1}{3}(e^x - 1)^1 + 1 \right] + C$
$I = 2\sqrt{e^x - 1} \left[ \frac{e^x - 1}{3} + \frac{3}{3} \right] + C$
$I = 2\sqrt{e^x - 1} \left[ \frac{e^x - 1 + 3}{3} \right] + C$
$I = 2\sqrt{e^x - 1} \left[ \frac{e^x + 2}{3} \right] + C$
$I = \frac{2}{3}(e^x + 2)\sqrt{e^x - 1} + C$

This is the simplified expression for the integral.



$$\begin{array}{rl}{\displaystyle \text{Let }u}& {\displaystyle =\sqrt{e^x-1}}\\ {\displaystyle u^2}& {\displaystyle =e^x-1}\\ {\displaystyle e^x}& {\displaystyle =u^2+1}\\ {\displaystyle e^{x}dx}& {\displaystyle =2udu}\\ {\displaystyle \int \frac{e^{2x}}{\sqrt{e^x-1}}dx}& {\displaystyle =\int \frac{e^x\cdot e^{x}dx}{\sqrt{e^x-1}}}\\ {\displaystyle }& {\displaystyle =\int \frac{(u^2+1)(2udu)}{u}}\\ {\displaystyle }& {\displaystyle =\int 2(u^2+1)du}\\ {\displaystyle }& {\displaystyle =\int (2u^2+2)du}\\ {\displaystyle }& {\displaystyle =2\frac{u^3}{3}+2u+C}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}(\sqrt{e^x-1}{)}^3+2\sqrt{e^x-1}+C}\\ {\displaystyle }& {\displaystyle =\frac{2}{3}(e^x-1{)}^{3\mathrm{/}2}+2(e^x-1{)}^{1\mathrm{/}2}+C}\\ {\displaystyle }& {\displaystyle =2(e^x-1{)}^{1\mathrm{/}2}[\frac{1}{3}(e^x-1)+1]+C}\\ {\displaystyle }& {\displaystyle =2\sqrt{e^x-1}\left[\frac{e^x-1+3}{3}\right]+C}\\ {\displaystyle }& {\displaystyle =\boxed{{\displaystyle \frac{2}{3}(e^x+2)\sqrt{e^x-1}+C}}}\end{array}$$
\begin{align*}
\text{Let } u &= \sqrt{e^x - 1} \\
u^2 &= e^x - 1 \\
e^x &= u^2 + 1 \\
e^x dx &= 2u du \\
\int \frac{e^{2x}}{\sqrt{e^x - 1}} dx &= \int \frac{e^x \cdot e^x dx}{\sqrt{e^x - 1}} \\
&= \int \frac{(u^2+1)(2u du)}{u} \\
&= \int 2(u^2+1) du \\
&= \int (2u^2 + 2) du \\
&= 2\frac{u^3}{3} + 2u + C \\
&= \frac{2}{3}(\sqrt{e^x - 1})^3 + 2\sqrt{e^x - 1} + C \\
&= \frac{2}{3}(e^x - 1)^{3/2} + 2(e^x - 1)^{1/2} + C \\
&= 2(e^x - 1)^{1/2} \left[ \frac{1}{3}(e^x - 1) + 1 \right] + C \\
&= 2\sqrt{e^x - 1} \left[ \frac{e^x - 1 + 3}{3} \right] + C \\
&= \boxed{\frac{2}{3}(e^x + 2)\sqrt{e^x - 1} + C}
\end{align*}