int frac{x^5 , dx}{(x^2+x+1)(x^6+1)(x^4-x^3+x | vedclass

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

Question

(B) Let the integral be $I = \int \frac{x^5 \, dx}{(x^2+x+1)(x^6+1)(x^4-x^3+x-1)}$.First,simplify the denominator terms:$(x^2+x+1)(x^4-x^3+x-1) = (x^2

Vedclass · Accepted Answer

$\frac{1}{12} \log_e \left| \frac{x^6-1}{x^6+1} \right| + c$

Vedclass · Answer

(B) Let the integral be $I = \int \frac{x^5 \, dx}{(x^2+x+1)(x^6+1)(x^4-x^3+x-1)}$.First,simplify the denominator terms:$(x^2+x+1)(x^4-x^3+x-1) = (x^2+x+1)(x^3(x-1) + 1(x-1)) = (x^2+x+1)(x^3+1)(x-1)$.Since $(x^2+x+1)(x-1) = x^3-1$,the product becomes $(x^3-1)(x^3+1) = x^6-1$.Thus,the integral simplifies to $I = \int \frac{x^5 \, dx}{(x^6+1)(x^6-1)}$.Substitute $t = x^6$,then $dt = 6x^5 \, dx$,so $x^5 \, dx = \frac{1}{6} \, dt$.$I = \frac{1}{6} \int \frac{dt}{(t+1)(t-1)}$.Using partial fractions,$\frac{1}{(t+1)(t-1)} = \frac{1}{2} \left( \frac{1}{t-1} - \frac{1}{t+1} \right)$.$I = \frac{1}{6} \cdot \frac{1}{2} \int \left( \frac{1}{t-1} - \frac{1}{t+1} \right) dt = \frac{1}{12} (\log_e |t-1| - \log_e |t+1|) + c$.$I = \frac{1}{12} \log_e \left| \frac{t-1}{t+1} \right| + c = \frac{1}{12} \log_e \left| \frac{x^6-1}{x^6+1} \right| + c$.

$\int \frac{x^5 \, dx}{(x^2+x+1)(x^6+1)(x^4-x^3+x-1)} =$

Similar Questions

Integrate the rational function: $\frac{x}{(x^{2}+1)(x-1)}$

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

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

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

If $\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^2 \theta} d \theta=A \log _e|f(\theta)|+c$ (where $c$ is a constant of integration),then $\frac{f(\theta)}{A}$ can be

Vedclass Test Series

Exam Paper Generator

Online Exam Module