Question

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

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

Answer 1

The correct option is D $\frac{1}{10}{(4+\frac{1}{x^2})}^{-5}+c$

$I=\int \frac{x^{9}dx}{(4x^2+4{)}^6}$
$=\int \frac{dx}{x^3{(4+\frac{1}{x^2})}^6}$
Put $4+\frac{1}{x^2}=t$
$\Rightarrow -\frac{2}{x^3}dx=dt$
$=-\frac{1}{2}\int \frac{dt}{t^6}$
$=-\frac{1}{2}\frac{t^{-5}}{-5}+C$
$=\frac{1}{10}{(4+\frac{1}{x^2})}^{-5}+C$