Question

Find the integration of

Answer 1

Hey there,

Answer:
∫dx(x2+2x+2)2=12(tan−1(x+1)+x+1x2+2x+2)+c
Explanation:

∫dx(x2+2x+2)2=∫dx((x+1)2+1)2

Let x+1=tanu⇒dx=sec2udu

∫dx((x+1)2+1)2=∫sec2u(1+tan2u)2du

=∫sec2u(sec2u)2du

=∫1sec2udu

=∫cos2udu

=∫12(1+cos2u)du

=12∫(1+cos2u)du

=12(u+12sin2u)+c

Since tanu=x+1⇒sinu=x+1√x2+2x+2andcosu=1√x2+2x+2

=12(tan−1(x+1)+122sinucosu)+c

=12(tan−1(x+1)+x+1√x2+2x+2⋅1√x2+2x+2)+c

=12(tan−1(x+1)+x+1x2+2x+2)+c

Hope this helps!