$\int_{-1}^{1} \dfrac{x^2}{1+x^2}\,dx$ equals:
- $2 - \pi/4$
- $(4-\pi)/2$
- $\pi/4$
- $0$
Show answer and solution
Correct answer: B. $(4-\pi)/2$
Integrand $x^2/(1+x^2)$ is **even**, so $I = 2\int_0^1 x^2/(1+x^2)\,dx = 2\int_0^1 [1 - 1/(1+x^2)]\,dx$ $= 2[x - \tan^{-1}x]_0^1 = 2[1 - \pi/4] = (4-\pi)/2$.