Home › MHT-CET › Mathematics › Definite Integration › $\int_{-1}^{1} \dfrac{x^2}{1+x^2}\,dx$ equals:
$\int_{-1}^{1} \dfrac{x^2}{1+x^2}\,dx$ 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 $I = 2\int_0^1 x^2/(1+x^2)\,dx = 2\int_0^1 [1 - 1/(1+x^2)]\,dx$ $= 2[x - \tan^{-1}x]_0^1 = 2[1 - \pi/4] = (4-\pi)/2$.
Related questions
Using definite integration as the limit of a sum, $\int_1^2 (2x + 5)\,dx$ equals:$\int_0^{\pi/2} \cos^2 x\,dx$ equals:$\int_0^{\pi/2} qrt{1 - \cos 4x}\,dx$ equals:Using the King's property, evaluate $\int_0^{\pi/2} \dfrac{dx}{1 + qrt[3]{\tan x}}$:$\int_1^2 \dfrac{\log x}{x^2}\,dx$ equals:$\int_{-1}^{1} |5x - 3|\,dx$ equals:Evaluate $\int_2^3 7^x\,dx$:$\int_0^4 (x - x^2)\,dx$ equals: