The evaluation of int {frac{{p{x^{p + 2q - 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}}

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 ${x^{2p + 2q}}$: $I = \int {\frac{{p{x^{p + 2q - 1}} - q{x^{q - 1}}}}{{{x^{2p + 2q}} (1 + {x^{-(p+q)}})^2}}} \,dx$ $I = \int {\frac{{p{x^{-p-1}} - q{x^{-p-q-1}}}}{{{{(1 + {x^{-(p+q)}})}^2}}}} \,dx$. Let $u = 1 + {x^{-(p+q)}}$. Then $du = -(p+q) {x^{-(p+q)-1}} \,dx$. This approach is complex. Alternatively,divide numerator and denominator by ${x^{2q}}$: $I = \int {\frac{{p{x^{p-1}} - q{x^{-q-1}}}}{{{{({x^p} + {x^{-q}})}^2}}}} \,dx$. Let $t = {x^p} + {x^{-q}}$. Then $dt = (p{x^{p-1}} - q{x^{-q-1}}) \,dx$. Thus,$I = \int {t^{-2}} \,dt = -{t^{-1}} + C = -\frac{1}{{{x^p} + {x^{-q}}}} + C$. Multiply numerator and denominator by ${x^q}$: $I = -\frac{{{x^q}}}{{{x^{p+q}} + 1}} + C$.

The evaluation of $\int {\frac{{p{x^{p + 2q - 1}} - q{x^{q - 1}}}}{{{x^{2p + 2q}} + 2{x^{p + q}} + 1}}} \,dx$ is

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