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

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(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 \

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$(1, -2/7)$

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(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) = $

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