Section 2.5 Practice Gateway Exam 5
The use of calculators or any computer algebra system is not permitted on this test. No partial credit will be awarded. A score of at least 7 correct is required to pass this gateway.
Compute each of the following integrals.
Exercises Exercises
1.
\(\displaystyle \int \frac{(\ln(x))^3}{x} \, dx \)
Solution.
Use substitution with \(u= \ln(x)\) and \(du = \dfrac{1}{x} \, dx\text{:}\)
\begin{align*}
\int \frac{(\ln(x))^3}{x} \, dx \amp = \int u^3 \, du\\
\amp = \frac{u^4}{4} + C\\
\amp = \frac{(\ln(x))^4}{4} + C
\end{align*}
2.
\(\displaystyle \int t^2e^t \, dt \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = t^2 \amp dv \amp = e^{t} \, dt\\
du \amp = 2t \, dt \amp v \amp = e^{t}
\end{align*}
Then
\begin{align*}
\int t^2e^{t} \, dt \amp = t^2\cdot e^{t} - \int e^{t} \cdot 2t \, dt\\
\amp =t^2 e^{t} - \int 2te^{t} \, dt
\end{align*}
We now use integration by parts again on the new integral.
\begin{align*}
u \amp = 2t \amp dv \amp = e^{t} \, dt\\
du \amp = 2\, dt \amp v \amp = e^{t}
\end{align*}
So
\begin{align*}
\int t^2e^{t} \, dt \amp = t^2 e^{t} - \left( 2t e^{t} - \int 2e^{t} \, dt\right)\\
\amp = t^2 e^{t} - 2te^{t} +2e^{t} + C
\end{align*}
3.
\(\displaystyle \int \frac{x^2+x}{x^2} \, dx \)
Solution.
\begin{align*}
\int \frac{x^2+x}{x^2} \, dx \amp = \int (1 + x^{-1}) \, dx\\
\amp = x + \ln|x| + C
\end{align*}
4.
\(\displaystyle \int \frac{3}{(x+2)(x+3)} \, dx \)
Solution.
First decompose the integrand into partial fractions:
\begin{equation*}
\frac{3}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}
\end{equation*}
This yields
\begin{equation*}
3 = A(x+3) + B(x+2)\text{.}
\end{equation*}
Plug in \(x=-2\) to solve for \(A\) and \(x=-3\) to solve for \(B\text{:}\)
\begin{equation*}
3 = A + 0 \implies A=3
\end{equation*}
\begin{equation*}
3 = 0 - B \implies B=-3
\end{equation*}
Now perform the integration:
\begin{align*}
\int \frac{3}{(x+2)(x+3)} \amp = \int\left(\frac{3}{x+2} - \frac{3}{x+3}\right) \, dx\\
\amp = 3\ln|x+2| - 3 \ln | x+3| + C
\end{align*}
5.
\(\displaystyle \int \sin^3(\theta)\cos(\theta) \, d\theta \)
Solution.
Use substitution with \(u= \sin(\theta)\) and \(du = \cos(\theta) \, d\theta\text{:}\)
\begin{align*}
\int \sin^3(\theta)\cos(\theta) \, d\theta \amp = \int u^3 \, du\\
\amp = \frac{u^4}{4} + C\\
\amp = \frac{\sin^4(\theta)}{4} + C
\end{align*}
6.
\(\displaystyle \int x\sin(4x) \, dx \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = x \amp dv \amp = \sin(4x) \, dx\\
du \amp = 1 \, dx \amp v \amp = -\frac{1}{4}\cos(4x)
\end{align*}
Then
\begin{align*}
\int x\sin(4x) \, dx \amp = x\cdot -\frac{1}{4}\cos(4x) - \int - \frac{1}{4} \cos(4x) \, dx\\
\amp = -\frac{1}{4}x\cos(4x) + \frac{1}{4} \int \cos(4x) \, dx\\
\amp = -\frac{1}{4}x\cos(4x) + \frac{1}{4} \cdot \frac{1}{4}\sin(4x) + C
\end{align*}
7.
\(\displaystyle \int \frac{x}{\sqrt{16+x^2}} \, dx \)
Solution.
Use substitution with \(u= 16+x^2\) and \(du = 2x \, dx\text{,}\) so \(\dfrac{1}{2} \, du = x \, dx\text{:}\)
\begin{align*}
\int \frac{x}{\sqrt{16+x^2}} \, dx \amp = \int \frac{1}{2} \cdot \frac{1}{\sqrt{u}} \, du\\
\amp = \frac{1}{2} \int u^{-1/2} \, du\\
\amp = \frac{1}{2} \cdot \frac{u^{1/2}}{1/2} + C\\
\amp = \sqrt{16+x^2} + C
\end{align*}
8.
\(\displaystyle \int \frac{e^z}{e^z-7} \, dz \)
Solution.
Use substitution with \(u=e^z-7\) and \(du = e^z \, dz\text{:}\)
\begin{align*}
\int \frac{e^z}{e^z-7} \, dz \amp = \int \frac{1}{u} \, du\\
\amp = \ln|u| + C\\
\amp = \ln|e^z - 7| + C
\end{align*}
