Evaluate the integral if it exists. $\int\left(\frac{5-x}{x}\right)^{2} d x$
A) $1-10 \ln x+C$
B) $x-10 \ln x-\frac{25}{x}+C$
C) $\ln x-25 x+C$
D) $x-\frac{1}{10 \ln x}+C$
E) none of these

Question

Quizplus · Accepted Answer

Use substitution $u=\frac{5-x}{x}$. Then $du=-\frac{6}{x^2}dx$. 
The integral becomes:
\begin{align*}
\int\left(\frac{5-x}{x}ight)^{2} d x &= \int \frac{(5-x)^2}{x^2} dx \
&= \int \left(\frac{25-10x+x^2}{x^2}ight)dx \
&= \int \left(\frac{u-10}{u^2}ight) (-\frac{5}{u^2})du \qquad\qquad \qquad	ext{Use } u=\frac{5-x}{x}\
&= \int \frac{(10-u)}{u^4}du \
&= 10\int u^{-4} du - \int u^{-3}du \
&= -\frac{10}{3u^3} + \frac{1}{2u^2} + C\
&= \frac{-10}{3(\frac{5-x}{x})^3} + \frac{1}{2(\frac{5-x}{x})^2} + C \
&= \frac{-10x^3}{(5-x)^3} + \frac{x^2}{2(5-x)^2}+C \
\end{align*}

Therefore, the answer is $\boxed{	extbf{(B)}\ x-10 \ln x-\frac{25}{x}+C}$.

Evaluate the Integral If It Exists ∫(5−xx)2dx\int\left(\frac{5-x}{x}\right)^{2} d x∫(x5−x​)2dx A) 1−10ln⁡x+C1-10 \ln x+C1−10lnx+C

Evaluate the Integral If It Exists $\int\left(\frac{5-x}{x}\right)^{2} d x$ A) $1-10 \ln x+C$