$\text { Find the integral } \int \frac { x - 15 } { x ^ { 2 } + 1 } d x \text {. }$
A) $15 \ln \left( x ^ { 2 } + 1 \right) - \arctan ( x ) + C$
B) $\frac { 1 } { 15 \ln \left( x ^ { 2 } + 1 \right) } - 15 \arctan ( x ) + C$
C) $\frac { 1 } { 2 \ln \left( x ^ { 2 } + 1 \right) } + 15 \arctan ( x ) + C$
D) $\frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right) - 15 \arctan ( x ) + C$
E) $2 \ln \left( x ^ { 2 } + 1 \right) - 15 \arctan ( x ) + C$

Question

$\text { Find the integral } \int \frac { x - 15 } { x ^ { 2 } + 1 } d x \text {. }$ 
A) $15 \ln \left( x ^ { 2 } + 1 \right) - \arctan ( x ) + C$
B) $\frac { 1 } { 15 \ln \left( x ^ { 2 } + 1 \right) } - 15 \arctan ( x ) + C$
C) $\frac { 1 } { 2 \ln \left( x ^ { 2 } + 1 \right) } + 15 \arctan ( x ) + C$
D) $\frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right) - 15 \arctan ( x ) + C$
E) $2 \ln \left( x ^ { 2 } + 1 \right) - 15 \arctan ( x ) + C$

Quizplus · Accepted Answer

We can begin by using substitution, letting $u=x^2+1$ and $du=2xdx$, so that $xdx=\frac{1}{2}du$. Then the integral becomes $$\int\frac{x-15}{x^2+1}dx=\int\frac{1}{2}\cdot\frac{u-16}{u}du=\frac{1}{2}\left(\int\frac{u}{u}du-\int\frac{16}{u}du\right)=\frac{1}{2}\ln(u)-8\ln|u|+C.$$ Substituting back in for $u$, we get $$\frac{1}{2}\ln(x^2+1)-8\ln(x^2+1)+C=\frac{1}{2}\ln\frac{x^2+1}{(x^2+1)^8}+C.$$ Using the fact that $\ln a^n=n\ln a$, we can simplify this to $$\frac{1}{2}\ln\frac{1}{(x^2+1)^7}+C=\boxed{\frac{1}{2}\ln(x^2+1)-\frac{7}{2}\ln|x^2+1|+C.}$$

Find the integral ∫x−15x2+1dx. \text { Find the integral } \int \frac { x - 15 } { x ^ { 2 } + 1 } d x \text {. } Find the integral ∫x2+1x−15​dx.

$\text { Find the integral } \int \frac { x - 15 } { x ^ { 2 } + 1 } d x \text {. }$