Evaluate the following Integrals:
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 
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