int frac{d x}{(x+a)^{frac{9}{7}}(x-b)^{frac{5 | vedclass

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Question

(C) Let $I = \int \frac{dx}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}}$.Rewrite the integrand as: $I = \int \frac{dx}{(x+a)^{\frac{9}{7}} \cdot (x+a)^{\f

Vedclass · Accepted Answer

$\frac{7}{2(a+b)}\left(\frac{x-b}{x+a}\right)^{\frac{2}{7}}+c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}}$.Rewrite the integrand as: $I = \int \frac{dx}{(x+a)^{\frac{9}{7}} \cdot (x+a)^{\frac{5}{7}} \cdot \left(\frac{x-b}{x+a}\right)^{\frac{5}{7}}} = \int \frac{dx}{(x+a)^2 \left(\frac{x-b}{x+a}\right)^{\frac{5}{7}}}$.Let $t = \frac{x-b}{x+a}$. Then $dt = \frac{(x+a)(1) - (x-b)(1)}{(x+a)^2} dx = \frac{a+b}{(x+a)^2} dx$.Thus,$\frac{dx}{(x+a)^2} = \frac{dt}{a+b}$.Substituting these into the integral,we get: $I = \int \frac{1}{t^{\frac{5}{7}}} \cdot \frac{dt}{a+b} = \frac{1}{a+b} \int t^{-\frac{5}{7}} dt$.Integrating,we get: $I = \frac{1}{a+b} \cdot \frac{t^{-\frac{5}{7}+1}}{-\frac{5}{7}+1} + c = \frac{1}{a+b} \cdot \frac{t^{\frac{2}{7}}}{\frac{2}{7}} + c = \frac{7}{2(a+b)} \left(\frac{x-b}{x+a}\right)^{\frac{2}{7}} + c$.

$\int \frac{d x}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}} = ?$

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