int frac{dx}{x(x^7 + 1)} = | vedclass

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

Question

(B) Given integral is $I = \int \frac{dx}{x(x^7 + 1)}$.Multiply the numerator and denominator by $x^6$ to get $I = \int \frac{x^6 dx}{x^7(x^7 + 1)}$.L

Vedclass · Accepted Answer

$\frac{1}{7}\log \left( \frac{x^7}{x^7 + 1} \right) + c$

Vedclass · Answer

(B) Given integral is $I = \int \frac{dx}{x(x^7 + 1)}$.Multiply the numerator and denominator by $x^6$ to get $I = \int \frac{x^6 dx}{x^7(x^7 + 1)}$.Let $x^7 = u$,then $7x^6 dx = du$,which implies $x^6 dx = \frac{du}{7}$.Substituting these into the integral,we get $I = \int \frac{du/7}{u(u + 1)} = \frac{1}{7} \int \frac{du}{u(u + 1)}$.Using partial fractions,$\frac{1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1}$.Thus,$I = \frac{1}{7} \int \left( \frac{1}{u} - \frac{1}{u + 1} \right) du = \frac{1}{7} (\log |u| - \log |u + 1|) + c$.$I = \frac{1}{7} \log \left| \frac{u}{u + 1} \right| + c$.Substituting $u = x^7$ back,we get $I = \frac{1}{7} \log \left( \frac{x^7}{x^7 + 1} \right) + c$.

$\int \frac{dx}{x(x^7 + 1)} = $

Similar Questions

$\int \frac{1 + \tan x}{x + \log \sec x} \, dx = $

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

$\int \frac{1+x \cos x}{x\left[1-x^2\left(e^{\sin x}\right)^2\right]} d x=$

The value of $\int \frac{1+x^{4}}{1+x^{6}} dx$ is

$\int \frac{2+\cos \frac{x}{2}}{x+\sin \frac{x}{2}} \,d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module