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

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

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

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

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