int frac{1}{[{(x - 1)}^3 {(x + 2)}^5]^{1/4}} | vedclass

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(A) Let $I = \int \frac{1}{[{(x - 1)}^3 {(x + 2)}^5]^{1/4}} \,dx$. We can rewrite the integrand as: $I = \int \frac{1}{[{(x - 1)}^3 {(x + 2)}^3 {(x +

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$\frac{4}{3}{\left( {\frac{{x - 1}}{{x + 2}}} \right)^{1/4}} + c$

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(A) Let $I = \int \frac{1}{[{(x - 1)}^3 {(x + 2)}^5]^{1/4}} \,dx$. We can rewrite the integrand as: $I = \int \frac{1}{[{(x - 1)}^3 {(x + 2)}^3 {(x + 2)}^2]^{1/4}} \,dx = \int \frac{1}{[{(x - 1)}^3 {(x + 2)}^3]^{1/4} \cdot {(x + 2)}^{2/4}} \,dx$ $I = \int \frac{1}{[\frac{x - 1}{x + 2}]^{3/4} \cdot {(x + 2)}^2} \,dx$. Let $t = \frac{x - 1}{x + 2}$. Then,by the quotient rule,$dt = \frac{(x + 2)(1) - (x - 1)(1)}{{(x + 2)}^2} \,dx = \frac{3}{{(x + 2)}^2} \,dx$. Thus,$\frac{1}{{(x + 2)}^2} \,dx = \frac{1}{3} \,dt$. Substituting these into the integral: $I = \int \frac{1}{t^{3/4}} \cdot \frac{1}{3} \,dt = \frac{1}{3} \int t^{-3/4} \,dt$. $I = \frac{1}{3} \left( \frac{t^{1/4}}{1/4} \right) + c = \frac{4}{3} t^{1/4} + c$. Substituting $t = \frac{x - 1}{x + 2}$ back,we get: $I = \frac{4}{3} {\left( \frac{x - 1}{x + 2} \right)}^{1/4} + c$.

$\int \frac{1}{[{(x - 1)}^3 {(x + 2)}^5]^{1/4}} \,dx$ is equal to

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