Integrate the function: frac{3x^{2}}{x^{6}+1} | vedclass

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

Question

(A) Let $x^{3} = t$. Then,differentiating both sides with respect to $x$,we get $3x^{2} dx = dt$. Substituting these into the integral,we have: $\int

Vedclass · Accepted Answer

$	an^{-1}(x^{3}) + C$

Vedclass · Answer

(A) Let $x^{3} = t$. Then,differentiating both sides with respect to $x$,we get $3x^{2} dx = dt$. Substituting these into the integral,we have: $\int \frac{3x^{2}}{x^{6}+1} dx = \int \frac{dt}{t^{2}+1}$. Using the standard integral formula $\int \frac{1}{x^{2}+1} dx = 	an^{-1}(x) + C$,we get: $= 	an^{-1}(t) + C$. Substituting back $t = x^{3}$,the final result is: $= 	an^{-1}(x^{3}) + C$,where $C$ is an arbitrary constant.

Integrate the function: $\frac{3x^{2}}{x^{6}+1}$

Similar Questions

The integral $\int \frac{(x^8-x^2) dx}{(x^{12}+3x^6+1) \tan^{-1}(x^3+\frac{1}{x^3})}$ is equal to:

$\int \sec^p x \tan x \, dx = $

$\int \sin \sqrt{x} \,d x=\ldots+C$ (where $C$ is a constant of integration.)

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

Integrate the function $\frac{x}{\sqrt{x+4}}, x > 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module