# 009B Sample Final 1, Problem 4

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Compute the following integrals.

a) ${\displaystyle \int e^{x}(x+\sin(e^{x}))~dx}$

b) ${\displaystyle \int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}$

c) ${\displaystyle \int \sin ^{3}x~dx}$

Foundations:
Recall:
1. Integration by parts tells us that ${\displaystyle \int u~dv=uv-\int v~du}$.
2. Through partial fraction decomposition, we can write the fraction  ${\displaystyle {\frac {1}{(x+1)(x+2)}}={\frac {A}{x+1}}+{\frac {B}{x+2}}}$  for some constants ${\displaystyle A,B}$.
3. We have the Pythagorean identity ${\displaystyle \sin ^{2}(x)=1-\cos ^{2}(x)}$.

Solution:

(a)

Step 1:
We first distribute to get
${\displaystyle \int e^{x}(x+\sin(e^{x}))~dx\,=\,\int e^{x}x~dx+\int e^{x}\sin(e^{x})~dx.}$
Now, for the first integral on the right hand side of the last equation, we use integration by parts.
Let ${\displaystyle u=x}$ and ${\displaystyle dv=e^{x}dx}$. Then, ${\displaystyle du=dx}$ and ${\displaystyle v=e^{x}}$.
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int e^{x}(x+\sin(e^{x}))~dx}&=&\displaystyle {{\bigg (}xe^{x}-\int e^{x}~dx{\bigg )}+\int e^{x}\sin(e^{x})~dx}\\&&\\&=&\displaystyle {xe^{x}-e^{x}+\int e^{x}\sin(e^{x})~dx}.\\\end{array}}}$
Step 2:
Now, for the one remaining integral, we use ${\displaystyle u}$-substitution.
Let ${\displaystyle u=e^{x}}$. Then, ${\displaystyle du=e^{x}dx}$.
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int e^{x}(x+\sin(e^{x}))~dx}&=&\displaystyle {xe^{x}-e^{x}+\int \sin(u)~du}\\&&\\&=&\displaystyle {xe^{x}-e^{x}-\cos(u)+C}\\&&\\&=&\displaystyle {xe^{x}-e^{x}-\cos(e^{x})+C}.\\\end{array}}}$

(b)

Step 1:
First, we add and subtract ${\displaystyle x}$ from the numerator.
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {\int {\frac {2x^{2}+x-x+1}{2x^{2}+x}}~dx}\\&&\\&=&\displaystyle {\int {\frac {2x^{2}+x}{2x^{2}+x}}~dx+\int {\frac {1-x}{2x^{2}+x}}~dx}\\&&\\&=&\displaystyle {\int ~dx+\int {\frac {1-x}{2x^{2}+x}}~dx}.\\\end{array}}}$
Step 2:
Now, we need to use partial fraction decomposition for the second integral.
Since ${\displaystyle 2x^{2}+x=x(2x+1)}$, we let ${\displaystyle {\frac {1-x}{2x^{2}+x}}={\frac {A}{x}}+{\frac {B}{2x+1}}}$.
Multiplying both sides of the last equation by ${\displaystyle x(2x+1)}$,
we get ${\displaystyle 1-x=A(2x+1)+Bx}$.
If we let ${\displaystyle x=0}$, the last equation becomes ${\displaystyle 1=A}$.
If we let ${\displaystyle x=-{\frac {1}{2}}}$, then we get  ${\displaystyle {\frac {3}{2}}=-{\frac {1}{2}}\,B}$. Thus, ${\displaystyle B=-3}$.
So, in summation, we have  ${\displaystyle {\frac {1-x}{2x^{2}+x}}={\frac {1}{x}}+{\frac {-3}{2x+1}}}$.
Step 3:
If we plug in the last equation from Step 2 into our final integral in Step 1, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {\int ~dx+\int {\frac {1}{x}}~dx+\int {\frac {-3}{2x+1}}~dx}\\&&\\&=&\displaystyle {x+\ln x+\int {\frac {-3}{2x+1}}~dx}.\\\end{array}}}$
Step 4:
For the final remaining integral, we use ${\displaystyle u}$-substitution.
Let ${\displaystyle u=2x+1}$. Then, ${\displaystyle du=2\,dx}$ and  ${\displaystyle {\frac {du}{2}}=dx}$.
Thus, our final integral becomes
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx}&=&\displaystyle {x+\ln x+\int {\frac {-3}{2x+1}}~dx}\\&&\\&=&\displaystyle {x+\ln x+\int {\frac {-3}{2u}}~du}\\&&\\&=&\displaystyle {x+\ln x-{\frac {3}{2}}\ln u+C}.\\\end{array}}}$
${\displaystyle \int {\frac {2x^{2}+1}{2x^{2}+x}}~dx\,=\,x+\ln x-{\frac {3}{2}}\ln(2x+1)+C.}$

(c)

Step 1:
First, we write ${\displaystyle \int \sin ^{3}x~dx=\int \sin ^{2}x\sin x~dx}$.
Using the identity ${\displaystyle \sin ^{2}x+\cos ^{2}x=1}$, we get ${\displaystyle \sin ^{2}x=1-\cos ^{2}x}$.
If we use this identity, we have
${\displaystyle \int \sin ^{3}x~dx=\int (1-\cos ^{2}x)\sin x~dx}$.
Step 2:
Now, we proceed by ${\displaystyle u}$-substitution. Let ${\displaystyle u=\cos x}$. Then, ${\displaystyle du=-\sin xdx}$.
So we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int \sin ^{3}x~dx}&=&\displaystyle {\int -(1-u^{2})~du}\\&&\\&=&\displaystyle {-u+{\frac {u^{3}}{3}}+C}\\&&\\&=&\displaystyle {-\cos x+{\frac {\cos ^{3}x}{3}}+C}.\\\end{array}}}$
(a)  ${\displaystyle xe^{x}-e^{x}-\cos(e^{x})+C}$
(b)  ${\displaystyle x+\ln x-{\frac {3}{2}}\ln(2x+1)+C}$
(c)  ${\displaystyle -\cos x+{\frac {\cos ^{3}x}{3}}+C}$