Question

$\int \frac{x^4+1}{x^6+1}dx={\tan }^{1}x+\frac{1}{k}{\tan }^{-1}{x}^3$. Find the value of $k$.

Answer 1

The factors of $x^6+1$ are $(x^2+1)(x^4-x^2+1)$
Add and subtract $x^2$ in the $N^r.$
$\therefore I=\int \frac{x^4-x^2+1+x^2}{x^6+1}dx$
$=\int \frac{1}{x^2+1}dx+\frac{1}{3}\int \frac{3x^2}{{\left(x^3\right)}^2+1}dx$
$={\tan }^{1}x+\frac{1}{3}{\tan }^{-1}{x}^3.$