allery$\int_{-1}^{1} \dfrac{x^2}{1+x^2}\,dx$ equals:
A$2 - \pi/4$
B$(4-\pi)/2$
C$\pi/4$
D$0$
Answer & Solution
Correct answer: B. $(4-\pi)/2$
Integrand $x^2/(1+x^2)$ is **even**, so $I = 2\int_0^1 x^2/(1+x^2)\,dx = 2\int_0^1 [1 - 1/(1+x^2)]\,dx$ $= 2[x - \tan^{-1}x]_0^1 = 2[1 - \pi/4] = (4-\pi)/2$.
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