Question

Evaluate $\int \frac{x^2}{9+16x^6}dx$

• A. $\frac{1}{16}{\tan }^{-1}\left(\frac{4x^3}{3}\right)+c$
• B. $\frac{1}{36}{\tan }^{-1}\left(\frac{3x^3}{4}\right)+c$
• C. $\frac{1}{16}{\tan }^{-1}\left(\frac{3x^3}{4}\right)+c$
• D. $\frac{1}{36}{\tan }^{-1}\left(\frac{4x^3}{3}\right)+c$

Answer 1

The correct option is D $\frac{1}{36}{\tan }^{-1}\left(\frac{4x^3}{3}\right)+c$

$I=\int \frac{x^2}{9+16x^6}dx$
Substitute $x^3=t=>3x^{2}dx=dt$
$\Rightarrow I=\frac{1}{3}\int \frac{1}{9+16t^2}dt=\frac{1}{48}\int \frac{1}{(3/4{)}^2+t^2}dt$
$=\frac{1}{48}.\frac{1}{3/4}{\tan }^{-1}\left(\frac{4t}{3}\right)+c$
$=\frac{1}{36}{\tan }^{-1}\left(\frac{4x^3}{3}\right)+c$