int frac{1 - x^7}{x(1 + x^7)} dx equals: | vedclass

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(C) Let $I = \int \frac{1 - x^7}{x(1 + x^7)} dx$. We can rewrite the integrand as: $\frac{1 - x^7}{x(1 + x^7)} = \frac{1 + x^7 - 2x^7}{x(1 + x^7)} = \

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$\ln |x| - \frac{2}{7} \ln (1 + x^7) + c$

Vedclass · Answer

(C) Let $I = \int \frac{1 - x^7}{x(1 + x^7)} dx$. We can rewrite the integrand as: $\frac{1 - x^7}{x(1 + x^7)} = \frac{1 + x^7 - 2x^7}{x(1 + x^7)} = \frac{1}{x} - \frac{2x^7}{x(1 + x^7)} = \frac{1}{x} - \frac{2x^6}{1 + x^7}$. Now,integrate term by term: $I = \int \frac{1}{x} dx - \int \frac{2x^6}{1 + x^7} dx$. For the second integral,let $u = 1 + x^7$,then $du = 7x^6 dx$,which implies $x^6 dx = \frac{du}{7}$. $I = \ln |x| - 2 \int \frac{1}{u} \cdot \frac{du}{7} = \ln |x| - \frac{2}{7} \ln |u| + c$. Substituting $u = 1 + x^7$ back,we get: $I = \ln |x| - \frac{2}{7} \ln (1 + x^7) + c$.

$\int \frac{1 - x^7}{x(1 + x^7)} dx$ equals:

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