Question

$\int \frac{(t^2+1{)}^2}{t^6+1}dt$

Answer 1

Let,

$I=\int \frac{{(t^2+1)}^2}{t^6}dt$

$=\int \frac{t^4+1+2t^2}{t^6}dt=\int (\frac{1}{t^2}+\frac{1}{t^4}+\frac{2}{t^6})dt$

$=\int t^{-2}dt+\int t^{-4}dt+2\int t^{-6}dt$

$=\frac{t^{-1}}{-1}+\frac{t^{-3}}{-3}+\frac{2t^{-5}}{-5}+C$

$=\frac{1}{t}-\frac{1}{3t^3}-\frac{2}{5t^5}+C$

Hence, this is the answer.