int tan^{-1} left( frac{2x}{1 - x^2} right) d | vedclass

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

Question

(D) Let $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. Substituting these into the integral: $\int 	an^{-1} \left( \frac{2 	an 	heta}{1 -

Vedclass · Accepted Answer

$2x 	an^{-1} x - \log(1 + x^2) + c$

Vedclass · Answer

(D) Let $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. Substituting these into the integral: $\int 	an^{-1} \left( \frac{2 	an 	heta}{1 - 	an^2 	heta} ight) \sec^2 	heta \, d	heta = \int 	an^{-1}(	an 2	heta) \sec^2 	heta \, d	heta$ $= \int 2	heta \sec^2 	heta \, d	heta = 2 \int 	heta \sec^2 	heta \, d	heta$. Using integration by parts: $= 2 \left[ 	heta 	an 	heta - \int 	an 	heta \, d	heta ight] = 2 \left[ 	heta 	an 	heta - \log|\sec 	heta| ight] + c$ $= 2 	heta 	an 	heta - 2 \log|\sec 	heta| + c$. Since $	an 	heta = x$,then $	heta = 	an^{-1} x$ and $\sec 	heta = \sqrt{1 + x^2}$. $= 2x 	an^{-1} x - 2 \log(\sqrt{1 + x^2}) + c$ $= 2x 	an^{-1} x - 2 \cdot \frac{1}{2} \log(1 + x^2) + c$ $= 2x 	an^{-1} x - \log(1 + x^2) + c$.

$\int \tan^{-1} \left( \frac{2x}{1 - x^2} \right) dx = $

Similar Questions

Evaluate the integral: $\int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x$

$\int x^4 e^{2 x} d x=$

$\int \log x \, dx = $

Let $\int x^3 \sin x \, dx = g(x) + C$,where $C$ is the constant of integration. If $8\left(g\left(\frac{\pi}{2}\right) + g^{\prime}\left(\frac{\pi}{2}\right)\right) = \alpha \pi^3 + \beta \pi^2 + \gamma$,where $\alpha, \beta, \gamma \in \mathbb{Z}$,then $\alpha + \beta - \gamma$ equals:

On $I \subset R-\{-1,1\}, \int \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module