int frac{1-x^7}{x(1+x^7)} dx = a ln |x| + b l | vedclass

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

Question

(C) We have the integral $I = \int \frac{1-x^7}{x(1+x^7)} dx$. Multiply the numerator and denominator by $x^6$ to simplify the expression: $I = \int \

Vedclass · Accepted Answer

$(1, -2/7)$

Vedclass · Answer

(C) We have the integral $I = \int \frac{1-x^7}{x(1+x^7)} dx$. Multiply the numerator and denominator by $x^6$ to simplify the expression: $I = \int \frac{x^6(1-x^7)}{x^7(1+x^7)} dx$. Let $u = x^7$,then $du = 7x^6 dx$,which implies $x^6 dx = \frac{du}{7}$. Substituting these into the integral: $I = \int \frac{1-u}{u(1+u)} \cdot \frac{du}{7} = \frac{1}{7} \int \frac{1-u}{u(1+u)} du$. Using partial fractions: $\frac{1-u}{u(1+u)} = \frac{A}{u} + \frac{B}{1+u}$. $1-u = A(1+u) + Bu$. For $u=0$,$A=1$. For $u=-1$,$1-(-1) = B(-1) \Rightarrow B = -2$. So,$I = \frac{1}{7} \int (\frac{1}{u} - \frac{2}{1+u}) du = \frac{1}{7} (\ln |u| - 2 \ln |1+u|) + C$. Substituting $u = x^7$ back: $I = \frac{1}{7} \ln |x^7| - \frac{2}{7} \ln |x^7+1| + C = \frac{7}{7} \ln |x| - \frac{2}{7} \ln |x^7+1| + C = 1 \ln |x| - \frac{2}{7} \ln |x^7+1| + C$. Comparing this with $a \ln |x| + b \ln |x^7+1| + c$,we get $a = 1$ and $b = -2/7$.

$\int \frac{1-x^7}{x(1+x^7)} dx = a \ln |x| + b \ln |x^7+1| + c \Rightarrow (a, b) = $

Similar Questions

The integral $\int \frac{1}{\sqrt[4]{(x-1)^{3}(x+2)^{5}}} dx$ is equal to : (where $C$ is a constant of integration)

$\int {x \cos(x^2) \, dx}$ is equal to

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

$\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} d x=$

$\int \cos ^{3} x \cdot e^{\log (\sin x)} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module