Question

Solve:
$\int \frac{x^4+1}{x^6+1}dx$

Answer 1

We have,

\[\begin{align}

$I=\int \frac{x^4+1}{x^6+1}dx$

$I=\int (\frac{x^4+1}{x^6+1}\times \frac{x^2+1}{x^2+1})dx$

$I=\int \left(\frac{x^6+x^2+x^4+1}{(x^6+1)(x^2+1)}\right)dx$

$I=\int (\frac{x^6+1}{(x^6+1)(x^2+1)}\times \frac{x^2+x^4}{(x^6+1)(x^2+1)})dx$

$I=\int \frac{1}{x^2+1}dx+\frac{1}{3}\int \frac{3x^2}{x^6+1}dx$

$I=\int \frac{1}{x^2+1}dx+\frac{1}{3}\int \frac{3x^2}{{\left(x^3\right)}^2+1}dx$

Let

$x^3=t$

$3x^{2}dx=dt$

Then,

$I=\int \frac{1}{x^2+1}dx+\frac{1}{3}\int \frac{dt}{t^2+1}$

On integrating and we get,

$I={\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}t+C$

Put $t=x^3$ and we get,

$I={\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}{x}^3+C$

Hence, this is the answer.