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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