int frac{2x tan^{-1}(x^2)}{1 + x^4} , dx = | vedclass

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

Question

(B) Let $t = 	an^{-1}(x^2)$.Then,differentiating both sides with respect to $x$,we get $dt = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) \, dx = \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $t = 	an^{-1}(x^2)$.Then,differentiating both sides with respect to $x$,we get $dt = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) \, dx = \frac{2x}{1 + x^4} \, dx$.Substituting these into the integral,we have:$\int 	an^{-1}(x^2) \cdot \frac{2x}{1 + x^4} \, dx = \int t \, dt$.Integrating with respect to $t$,we get $\frac{t^2}{2} + c$.Substituting back the value of $t$,we obtain $\frac{1}{2} (	an^{-1}(x^2))^2 + c$.

$\int \frac{2x \tan^{-1}(x^2)}{1 + x^4} \, dx = $

Similar Questions

$\int {\frac{{\sec x(1 + \tan x)dx}}{{({e^{ - x}} + \sec x)}}} = f(x) + C$ where $f(0) = \ln 2$,then $f\left( {\frac{\pi }{4}} \right)$ is -

$\int \frac{1}{x^2\sqrt{1+x^2}} dx =$

$\int \frac{\sin 2x}{(a+b \cos x)^2} dx =$

$\int \frac{\sec x}{\sqrt{\sin (2 x + \theta) + \sin \theta}} d x =$

$\int \frac{d x}{\cos x \sqrt{\cos 2 x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module