Question

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

Answer 1

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

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

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

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

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

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

Put, $x^3=t$

$\Rightarrow 3x^{2}dx=dt$

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

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

$={\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}\left(t\right)+c$

$={\tan }^{-1}\left(x\right)+\frac{1}{3}{\tan }^{-1}\left(x^3\right)+c$.