009B Sample Final 1, Problem 4

Compute the following integrals.

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

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

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

Foundations:
Review $u$-substitution
Integration by parts
Partial fraction decomposition
Trig identities

Solution:

(a)

Step 1:
We first distribute to get $\int e^{x}(x+\sin(e^{x}))~dx=\int e^{x}x~dx+\int e^{x}\sin(e^{x})~dx$.
Now, for the first integral on the right hand side of the last equation, we use integration by parts.
Let $u=x$ and $dv=e^{x}dx$. Then, $du=dx$ and $v=e^{x}$. So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{x}(x+\sin(e^{x}))~dx={\bigg (}xe^{x}-\int e^{x}~dx{\bigg )}+\int e^{x}\sin(e^{x})~dx=xe^{x}-e^{x}+\int e^{x}\sin(e^{x})~dx}

Step 2:
Now, for the one remaining integral, we use $u$-substitution.
Let $u=e^{x}$. Then, $du=e^{x}dx$. So, we have
$\int e^{x}(x+\sin(e^{x}))~dx=xe^{x}-e^{x}+\int \sin(u)~du=xe^{x}-e^{x}-\cos(u)+C=xe^{x}-e^{x}-\cos(e^{x})+C$.

(b)

Step 1:
First, we add and subtract $x$ from the numerator. So, we have
$\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx=\int {\frac {2x^{2}+x-x+1}{2x^{2}+x}}~dx=\int {\frac {2x^{2}+x}{2x^{2}+x}}~dx+\int {\frac {1-x}{2x^{2}+x}}~dx=\int ~dx+\int {\frac {1-x}{2x^{2}+x}}~dx$.

Step 2:
Now, we need to use partial fraction decomposition for the second integral.
Since $2x^{2}+x=x(2x+1)$, we let ${\frac {1-x}{2x^{2}+x}}={\frac {A}{x}}+{\frac {B}{2x+1}}$.
Multiplying both sides of the last equation by $x(2x+1)$, we get $1-x=A(2x+1)+Bx$.
If we let $x=0$, the last equation becomes $1=A$.
If we let $x=-{\frac {1}{2}}$, then we get ${\frac {3}{2}}=-{\frac {1}{2}}B$. Thus, $B=-3$.
So, in summation, we have ${\frac {1-x}{2x^{2}+x}}={\frac {1}{x}}+{\frac {-3}{2x+1}}$.

Step 3:
If we plug in the last equation from Step 2 into our final integral in Step 1, we have
$\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx=\int ~dx+\int {\frac {1}{x}}~dx+\int {\frac {-3}{2x+1}}~dx=x+\ln x+\int {\frac {-3}{2x+1}}~dx$.

Step 4:
For the final remaining integral, we use $u$-substitution.
Let $u=2x+1$. Then, $du=2dx$ and ${\frac {du}{2}}=dx$.
Thus, our final integral becomes
$\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx=x+\ln x+\int {\frac {-3}{2x+1}}~dx=x+\ln x+\int {\frac {-3}{2u}}~du=x+\ln x-{\frac {3}{2}}\ln u+C$.
Therefore, the final answer is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\sin^3x~dx=\int \sin^2 x \sin x~dx} .
Using the identity Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin^2x+\cos^2x=1} , we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin^2x=1-\cos^2x} . If we use this identity, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx} .

Step 2:
Now, we proceed by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=\cos x} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=-\sin x dx} . So we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\sin^3x~dx=\int -(1-u^2)~du=-u+\frac{u^3}{3}+C=-\cos x+\frac{\cos^3x}{3}+C} .

Final Answer:
(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle xe^x-e^x-\cos(e^x)+C}
(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+\ln x-\frac{3}{2}\ln (2x+1) +C}
(c) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\cos x+\frac{\cos^3x}{3}+C}

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