Question

Integrate: $\int \frac{x^4+1}{x^2+1}dx$

Answer 1

$\int \frac{x^4+1}{x^2+1}dx=\int \frac{x^4-1+1+1}{x^2+1}dx$

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

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

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

$\Rightarrow =\int x^{2}dx-\int 1dx+\int \frac{1}{x^2+1}dx$

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

Hence $\int \frac{x^4+1}{x^2+1}dx=\frac{x^3}{3}-x+{\tan }^{-1}x+C$