int {frac{{1 - {x^7}}}{{x(1 + {x^7})}}} ,dx e | 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}) - 2{x^7}}{x

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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