find the anti - derivative and simplify.\n int\frac{x^{2}}{x^{6}+16}dx \n \frac{?^{-1}(\frac{x}{})}{}+c

Explanation:

Step1: Use substitution

Let $u = x^{3}$, then $du=3x^{2}dx$ and $x^{2}dx=\frac{1}{3}du$. The integral $\int\frac{x^{2}}{x^{6}+16}dx$ becomes $\frac{1}{3}\int\frac{du}{u^{2}+16}$.

Step2: Recall integral formula

We know that $\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\tan^{- 1}(\frac{x}{a})+C$. Here $a = 4$ and $x = u$, so $\frac{1}{3}\int\frac{du}{u^{2}+16}=\frac{1}{3}\times\frac{1}{4}\tan^{-1}(\frac{u}{4})+C$.

Step3: Substitute back

Substitute $u = x^{3}$ back into the expression, we get $\frac{1}{12}\tan^{-1}(\frac{x^{3}}{4})+C$.

Answer:

$\frac{1}{12}\tan^{-1}(\frac{x^{3}}{4})+C$