Question

$\int\frac{x^{9}}{\left(4x^{2} +1\right)^{6}} dx$ is equal to

• A. $\frac{1}{5x} \left(4+ \frac{1}{x^{2}}\right)^{-5} + C$
• B. $\frac{1}{5} \left(4+ \frac{1}{x^{2}}\right)^{-5} + C$
• C. $\frac{1}{10x} \left(1+4\right)^{-5} + C$
• D. $\frac{1}{10x} \left(\frac{1}{x^{2}}+4\right)^{-5} + C$

Answer 1

Answer: (D)

We have, $I = \int\frac{x^{9}}{\left(4x^{2} +1\right)^{6}} dx $

$ = \int \frac{x^{9}dx}{x^{12}\left(4 +\frac{1}{x^{2}}\right)^{6}} = \int \frac{dx}{x^{3}\left(4 +\frac{1}{x^{2}}\right)^{6}} $

Put $4 + \frac{1}{x^{2}} = t $

$\Rightarrow \frac{-2}{x^{3}} dx = dt $

$\therefore I= -\frac{1}{2} \int\frac{dt}{t^{6}} = \frac{-1}{2} \int t^{-6} dt $

$= -\frac{1}{2} \frac{t^{-5}}{-5} +C $

$= \frac{1}{10}\left(4+ \frac{1}{x^{2}} \right) ^{-5} +C$