The integral int frac{dx}{(x+4)^{frac{8}{7}}( | vedclass

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(A) Let $I = \int \frac{dx}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}}$.We can rewrite the integrand as $I = \int \frac{dx}{\left(\frac{x+4}{x-3}\right)^

Vedclass · Accepted Answer

$\left(\frac{x-3}{x+4}\right)^{\frac{1}{7}}+C$

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(A) Let $I = \int \frac{dx}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}}$.We can rewrite the integrand as $I = \int \frac{dx}{\left(\frac{x+4}{x-3}\right)^{\frac{8}{7}}(x-3)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}} = \int \frac{dx}{\left(\frac{x+4}{x-3}\right)^{\frac{8}{7}}(x-3)^{2}}$.Let $t = \frac{x+4}{x-3}$. Then $dt = \frac{(x-3)(1) - (x+4)(1)}{(x-3)^2} dx = \frac{-7}{(x-3)^2} dx$,which implies $\frac{dx}{(x-3)^2} = -\frac{1}{7} dt$.Substituting these into the integral,we get $I = \int \frac{1}{t^{\frac{8}{7}}} \left(-\frac{1}{7} dt\right) = -\frac{1}{7} \int t^{-\frac{8}{7}} dt$.Integrating,we get $I = -\frac{1}{7} \left( \frac{t^{-\frac{8}{7} + 1}}{-\frac{8}{7} + 1} \right) + C = -\frac{1}{7} \left( \frac{t^{-\frac{1}{7}}}{-\frac{1}{7}} \right) + C = t^{-\frac{1}{7}} + C$.Substituting $t = \frac{x+4}{x-3}$ back,we get $I = \left(\frac{x+4}{x-3}\right)^{-\frac{1}{7}} + C = \left(\frac{x-3}{x+4}\right)^{\frac{1}{7}} + C$.

The integral $\int \frac{dx}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}}$ is equal to (where $C$ is a constant of integration).

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