Question:
Mark the correct alternative in each of the following:
$\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$ is equal to
A. $\frac{1}{5 x}\left(4+\frac{1}{x^{2}}\right)^{-5}+C$
B. $\frac{1}{5}\left(4+\frac{1}{\mathrm{x}^{2}}\right)^{-5}+\mathrm{C}$
C. $\frac{1}{10 x}\left(\frac{1}{x^{2}}+4\right)^{-5}+C$
D. $\frac{1}{10}\left(\frac{1}{x^{2}}+4\right)^{-5}+C$
Solution:
$\mathrm{I}=\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$
$I=\int \frac{x^{9}}{x^{12}\left(4+\frac{1}{x^{2}} 6\right.} d x$
$I=\int \frac{1}{x^{3}\left(4+\frac{1}{x^{2}}\right)^{6}} d x$
Let $\left(4+\frac{1}{x^{2}}\right)=t ; \frac{-2}{x^{3}} d x=d t$
$I=\int \frac{d t}{-2 t^{6}}$
$I=\frac{1}{10}\left[\frac{1}{t^{5}}\right]$
$I=\frac{1}{10}\left(\left[4+\frac{1}{x^{2}}\right]^{-5}\right)+c$