Question

Solve:
$\int \frac{x^6+1}{x^2+1}dx$

Answer 1

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

Substitute $x=\tan t,dx={\sec }^{2}t.dt$

$\therefore I=\left(\frac{{\tan }^{6}t+1}{{\tan }^{2}t+1}\right){\sec }^{2}tdt$

$\therefore I=\int ({\tan }^{6}t+1)dt$

Now $\int {\tan }^{6}tdt=\int {\tan }^{4}t{\sec }^{2}tdt-\int {\tan }^{4}dt$

$=\frac{{\tan }^{5}t}{5}-\frac{{\tan }^{3}t}{3}+\int ({\sec }^{2}t-1)dt$

$\therefore \int {\tan }^{6}tdt=\frac{{\tan }^{5}t}{5}-\frac{{\tan }^{3}t}{3}+\tan t-t+c$

Substituting we get

$\int \frac{x^6+1}{x^2+1}dx=t+\frac{{\tan }^{5}t}{5}-\frac{{\tan }^{3}t}{3}+\tan t-t+c$

$\therefore \int \frac{x^6+1}{x^2+1}dx=\frac{x^5}{5}-\frac{x^3}{3}+x+c$

Ans : $\therefore \int \frac{x^6+1}{x^2+1}dx=\frac{x^5}{5}-\frac{x^3}{3}+x+c$