Question

Solve : $\int \frac{t^4-3t^2+2}{t^2(1+t^2)},dt$

Answer 1

$I=\int \frac{t^4-3t^2+2}{t^2(1+t^2)}dt$

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

$\Rightarrow \int dt-\int \frac{4t^2}{t^2(1+t^2)}dt+\int \frac{2}{t^2(1+t^2)}dt$

$I\Rightarrow t-\int \frac{4}{(1+t^2)}dt+\int 2[\frac{1}{t^2}-\frac{1}{1+t^2}]dt$

$I\Rightarrow t-4{\tan }^{-1}\left(t\right)+2\int \frac{1}{t^2}dt+\int \frac{2}{(1+t^2)}dt$

$I=t-4{\tan }^{-1}t+2\left(\frac{t^{-2+1}}{-2+1}\right)-2{\tan }^{-1}\left(t\right)+C$

$I\Rightarrow t-\frac{2}{t}-6{\tan }^{-1}\left(t\right)+C$

$\therefore \int \frac{t^4-3t^2+2}{t^2(1+t^2)}dt=t-\frac{2}{t}-6{\tan }^{-1}\left(t\right)+C$