int {frac{{p{x^{p + 2q - 1}} - q{x^{q - 1}}}} | vedclass

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Question

(C) Let $I = \int \frac{p x^{p+2q-1} - q x^{q-1}}{(x^{p+q} + 1)^2} dx$. Divide the numerator and denominator by $x^{2p+2q}$ or manipulate the expressi

Vedclass · Accepted Answer

$\frac{{ - {x^q}}}{{{x^{p + q}} + 1}} + C$

Vedclass · Answer

(C) Let $I = \int \frac{p x^{p+2q-1} - q x^{q-1}}{(x^{p+q} + 1)^2} dx$. Divide the numerator and denominator by $x^{2p+2q}$ or manipulate the expression: $I = \int \frac{p x^{p-1} - q x^{-q-1}}{(x^p + x^{-q})^2} dx$. Let $t = x^p + x^{-q}$. Then $dt = (p x^{p-1} - q x^{-q-1}) dx$. Substituting these into the integral,we get: $I = \int \frac{1}{t^2} dt = -\frac{1}{t} + C$. Substituting back $t = x^p + x^{-q} = \frac{x^{p+q} + 1}{x^q}$: $I = -\frac{1}{\frac{x^{p+q} + 1}{x^q}} + C = -\frac{x^q}{x^{p+q} + 1} + C$.

$\int {\frac{{p{x^{p + 2q - 1}} - q{x^{q - 1}}}}{{{x^{2p + 2q}} + 2{x^{p + q}} + 1}}dx} $ is equal to:

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