Q. 34

Evaluate the foll

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        Rate this question :

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