Question

integration x^4/x^2+1​

Answer 1

I=∫x2x4+1dx  

=12∫2x2x4+1dx

=12∫(x2+1)+(x2−1)x4+1dx

=12∫x2+1x4+1dx+12∫x2−1x4+1dx

=12∫1+1x2x2+1x2dx+12∫1−1x2x2+1x2dx

=12∫1+1x2(x−1x)2+2dx+12∫1−1x2(x2+1x)2−2dx

Let u=x−1x

du=(1+1x2)dx

v=x+1x

dv=(1−1x2)dx

I=12∫duu2+2+12∫dvv2−2

=12⋅12–√tan−1(u2–√)+12⋅122–√log∣∣∣v−2–√v+2–√∣∣∣+C

=122–√tan−1(x2−12–√x)+142–√log∣∣∣x2+1−2–√xx2+1+2–√x∣∣∣+C