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

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

Question

(A) To evaluate the integral $I = \int \frac{x}{1+x^4} \, dx$,we can use the method of substitution.Let $u = x^2$. Then,the derivative is $du = 2x \,

Vedclass · Accepted Answer

$\frac{1}{2} 	an^{-1}(x^2) + c$,where $c$ is the constant of integration

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{x}{1+x^4} \, dx$,we can use the method of substitution.Let $u = x^2$. Then,the derivative is $du = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, du$.Substituting these into the integral,we get:$I = \int \frac{1}{1+(x^2)^2} \cdot (x \, dx) = \int \frac{1}{1+u^2} \cdot \frac{1}{2} \, du$$I = \frac{1}{2} \int \frac{1}{1+u^2} \, du$Using the standard integral formula $\int \frac{1}{1+u^2} \, du = 	an^{-1}(u) + c$,we have:$I = \frac{1}{2} 	an^{-1}(u) + c$Substituting $u = x^2$ back into the expression,we get:$I = \frac{1}{2} 	an^{-1}(x^2) + c$Thus,the correct option is $A$.

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

Similar Questions

To find the value of $\int \frac{1 + \log x}{x} dx$,the proper substitution is

Integrate the function $\sqrt{\sin 2x} \cos 2x$.

Integrate the function: $\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}$

$\int \frac{dx}{x^2(x^4+1)^{3/4}}$ equals

$\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module