Question

Integration of [x / (x^2 +1)^2 ] dx

Answer 1

HELLO DEAR,

given function is ∫x/(x² + 1)².dx

Now,let (x² + 1) = t
⇒2x = dt/dx

⇒dx = dt/2x

So, ∫x/(x² + 1)² * dt/2x

⇒1/2 ∫dt/t²

⇒ 1/2∫ t^{-2}.dt

⇒1/2 (-1/t) + c

put the value of t in above function

⇒-1/2(x² + 1) + c


I HOPE ITS HELP YOU DEAR,
THANKS