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

If $f(x)$ and $g(x)$ are integrable functions, then which of the following is equal to $[\int f(x) dx][\int g(x) dx]$?

The positive integer $n \leq 5$ for which $\int_0^1 e^x(x-1)^n dx = 16-6e$ is

$\int \sin(\log x) \, dx = $

If $\int(\log x)^3 x^5 d x=\frac{x^6}{A}\left[B(\log x)^3+C(\log x)^2+D(\log x)-1\right]+k$ and $A, B, C, D$ are integers, then $A-(B+C+D)=$

If $\int f(x) dx = \psi(x)$,then $\int x^5 f(x^3) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module