We have,
∫dx(x2+1)√x2+1
So,
∫dx√(x2+1)2(x2+1)
⇒∫dx(x2+1)32
⇒∫(x2+1)−32dx
On integrating and we get,
(x2+1)−32−32+C
⇒−2(x2+1)−323+C
Hence, this is the answer.