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

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Evaluate the following Integral:

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

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

As we have the trigonometric identity 1 + tan2θ = sec2θ, to evaluate this integral we use x = tan θ

⇒ dx = sec2θ dθ (Differentiating both sides)

When x = 0, tan θ = 0 ⇒ θ = 0

When x = 1, tan θ = 1 ⇒ θ = $\cfrac{\pi}4$

So, the new limits are 0 and $\cfrac{\pi}4$.

Substituting this in the original integral,