int_{0}^{1} frac{x^{2}}{1+x^{2}} , dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_{0}^{1} \frac{x^{2}}{1+x^{2}} \, dx$. We can rewrite the integrand as $\frac{x^{2}+1-1}{1+x^{2}} = \frac{1+x^{2}}{1+x^{2}} - \frac{1

Vedclass · Accepted Answer

$1-\frac{\pi}{4}$

Vedclass · Answer

(B) Let $I = \int_{0}^{1} \frac{x^{2}}{1+x^{2}} \, dx$. We can rewrite the integrand as $\frac{x^{2}+1-1}{1+x^{2}} = \frac{1+x^{2}}{1+x^{2}} - \frac{1}{1+x^{2}} = 1 - \frac{1}{1+x^{2}}$. Now,integrate term by term: $I = \int_{0}^{1} \left( 1 - \frac{1}{1+x^{2}} ight) \, dx$. $I = \left[ x - 	an^{-1}(x) ight]_{0}^{1}$. Evaluating at the limits: $I = (1 - 	an^{-1}(1)) - (0 - 	an^{-1}(0))$. Since $	an^{-1}(1) = \frac{\pi}{4}$ and $	an^{-1}(0) = 0$,$I = 1 - \frac{\pi}{4} - 0 = 1 - \frac{\pi}{4}$.

$\int_{0}^{1} \frac{x^{2}}{1+x^{2}} \, dx =$

Similar Questions

$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx =$

$\int_0^\pi \frac{1}{4+3 \cos x} d x=$

If $I$ is the greatest of $I_1=\int_0^1 e^{-x} \cos ^2 x \, dx, I_2=\int_0^1 e^{-x^2} \cos ^2 x \, dx, I_3=\int_0^1 e^{-x^2} \, dx, I_4=\int_0^1 e^{-x^2 / 2} \, dx$, then

$\int_1^4 \left(x + \sqrt{x} + \frac{1}{x}\right) dx - \int_1^{2 \log 2} dx = $

$\int_0^\pi \sqrt{1+4 \sin^2 \frac{x}{2}+4 \sin \frac{x}{2}} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module