Answer :
Let 
Put 1 + x2 = t
⇒ 2xdx = dt (Differentiating both sides)
When x = 0, t = 1 + 02 = 1
When x = 1, t = 1 + 12 = 2
So, the new limits are 1 and 2.
In numerator, we can write 24x3dx = 12x2 × 2xdx
But, x2 = t – 1 and 2xdx = dt
⇒ 24x3dx = 12(t – 1)dt
Substituting this in the original integral,
Recall 
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