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