int frac{x}{1 + x^4} , dx = | vedclass

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

Question

(B) Let $t = x^2$. Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$. Substitutin

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $t = x^2$. Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$. Substituting these into the integral: $\int \frac{x}{1 + x^4} \, dx = \int \frac{1}{1 + (x^2)^2} \cdot (x \, dx) = \int \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt$. $= \frac{1}{2} \int \frac{1}{1 + t^2} \, dt$. Using the standard integral formula $\int \frac{1}{1 + t^2} \, dt = 	an^{-1}(t) + c$,we get: $= \frac{1}{2} 	an^{-1}(t) + c$. Substituting $t = x^2$ back,the final result is $\frac{1}{2} 	an^{-1}(x^2) + c$.

$\int \frac{x}{1 + x^4} \, dx = $

Similar Questions

$\int \frac{d x}{\left(e^x+e^{-x}\right)^2}=$

Evaluate the integral: $\int \frac{1}{\left(x+\frac{2}{x}\right) \sqrt{x^4+4 x^2+3}} d x$

$f(x) = \int \frac{dx}{\sin^6 x}$ is a polynomial of degree

If $\int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^{k}+c$,then the value of $k$ is

$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module