Evaluate the following Integral: ∫(24 x^3)/(1 + x^2)^4dx, x ∈ [0, 1]

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Evaluate the following Integral:

$\int\limits_0^1\cfrac{24\text x^3}{(1+\text x^2)^4}d\text x$

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Let I = $\int\limits_0^1\cfrac{24\text x^3}{(1+\text x^2)^4}d\text x$

Put 1 + x 2 = 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,