int frac{x^3 tan^{-1} x^4}{1+x^8} dx = | vedclass

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

Question

(A) Let $I = \int \frac{x^3 	an^{-1} x^4}{1+x^8} dx$. Substitute $t = x^4$,then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$. Substituting

Vedclass · Accepted Answer

$\frac{(	an^{-1}(x^4))^2}{8} + c$

Vedclass · Answer

(A) Let $I = \int \frac{x^3 	an^{-1} x^4}{1+x^8} dx$. Substitute $t = x^4$,then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$. Substituting these into the integral,we get $I = \frac{1}{4} \int \frac{	an^{-1} t}{1+t^2} dt$. Now,let $u = 	an^{-1} t$,then $du = \frac{1}{1+t^2} dt$. Substituting $u$ into the integral,we get $I = \frac{1}{4} \int u du$. Integrating with respect to $u$,we get $I = \frac{1}{4} \cdot \frac{u^2}{2} + c = \frac{u^2}{8} + c$. Substituting back $u = 	an^{-1} t$ and $t = x^4$,we get $I = \frac{(	an^{-1}(x^4))^2}{8} + c$.

$\int \frac{x^3 \tan^{-1} x^4}{1+x^8} dx =$

Similar Questions

Integrate the function $\frac{\sin x}{(1+\cos x)^{2}}$.

If $f(x) = \int \frac{1}{x^{1/4}(1+x^{1/4})} dx$ and $f(0) = -6$,then $f(1)$ is equal to:

$\int {{e^{{{\cos }^2}x}}\sin 2x\,dx = }$

$\int \frac{e^x \, dx}{\sqrt{1 - e^{2x}}} = $

$\int \frac{3x^2}{x^6 + 1} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module