If int frac{1}{sqrt[5]{(x-1)^4(x+3)^6}} dx = | vedclass

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(B) Given the integral $I = \int \frac{1}{(x-1)^{4/5}(x+3)^{6/5}} dx$. We can rewrite the integrand as: $I = \int \frac{1}{\left(\frac{x-1}{x+3}\right

Vedclass · Accepted Answer

$7$

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(B) Given the integral $I = \int \frac{1}{(x-1)^{4/5}(x+3)^{6/5}} dx$. We can rewrite the integrand as: $I = \int \frac{1}{\left(\frac{x-1}{x+3}ight)^{4/5} (x+3)^{4/5+6/5}} dx = \int \frac{1}{\left(\frac{x-1}{x+3}ight)^{4/5} (x+3)^2} dx$. Let $t = \frac{x-1}{x+3}$. Then $dt = \frac{(x+3)(1) - (x-1)(1)}{(x+3)^2} dx = \frac{4}{(x+3)^2} dx$. So,$\frac{1}{(x+3)^2} dx = \frac{1}{4} dt$. Substituting these into the integral: $I = \int \frac{1}{t^{4/5}} \cdot \frac{1}{4} dt = \frac{1}{4} \int t^{-4/5} dt = \frac{1}{4} \cdot \frac{t^{1/5}}{1/5} + C = \frac{5}{4} \left(\frac{x-1}{x+3}ight)^{1/5} + C$. Comparing this with $A\left(\frac{\alpha x-1}{\beta x+3}ight)^B + C$,we get $A = \frac{5}{4}$,$\alpha = 1$,$\beta = 1$,and $B = \frac{1}{5}$. Finally,calculating $\alpha + \beta + 20AB = 1 + 1 + 20 	imes \frac{5}{4} 	imes \frac{1}{5} = 2 + 5 = 7$.

If $\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} dx = A\left(\frac{\alpha x-1}{\beta x+3}\right)^B + C,$ where $C$ is the constant of integration,then the value of $\alpha + \beta + 20AB$ is...........

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