int_1^2 left( tan ^{-1}left(frac{x}{x^2+1}rig | vedclass

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

Question

(D) We know that for $y > 0$,$	an ^{-1}(y) + 	an ^{-1}(1/y) = \frac{\pi}{2}$.Let $y = \frac{x}{x^2+1}$. Since $x \in [1, 2]$,$y > 0$.Therefore,the i

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) We know that for $y > 0$,$	an ^{-1}(y) + 	an ^{-1}(1/y) = \frac{\pi}{2}$.Let $y = \frac{x}{x^2+1}$. Since $x \in [1, 2]$,$y > 0$.Therefore,the integrand simplifies as follows:$	an ^{-1}\left(\frac{x}{x^2+1}ight) + 	an ^{-1}\left(\frac{x^2+1}{x}ight) = \frac{\pi}{2}$.Now,evaluate the integral:$I = \int_1^2 \frac{\pi}{2} d x$$I = \frac{\pi}{2} [x]_1^2$$I = \frac{\pi}{2} (2 - 1) = \frac{\pi}{2}$.

$\int_1^2 \left( \tan ^{-1}\left(\frac{x}{x^2+1}\right)+\tan ^{-1}\left(\frac{x^2+1}{x}\right) \right) d x =$

Similar Questions

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$, then $b$ is equal to

If $I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x$,where $n$ is a positive integer,then $I_{10}+I_{8}$ is

Evaluate the definite integral $\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$.

$\int_{1}^{3} (x - 1)(x - 2)(x - 3) \, dx = $

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\int_{\pi /2}^x {t\,dt} }}{{\sin (2x - \pi )}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module