Let

Put x⁵ + 1 = t

⇒ 5x⁴dx = dt (Differentiating both sides)

When x = –1, t = (–1)⁵ + 1 = 0

When x = 1, t = 1⁵ + 1 = 2

So, the new limits are 0 and 2.

Substituting this in the original integral,

Recall