If int frac{1}{left((x+4)^3(x+1)^5right)^{1 / | vedclass

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Question

(B) Given the integral $I = \int \frac{1}{\left((x+4)^3(x+1)^5\right)^{1 / 4}} d x$. Rewrite the integrand as $I = \int \frac{1}{\left((x+4)^3(x+1)^3(

Vedclass · Accepted Answer

$n + \frac{1}{A} = -\frac{1}{2}$

Vedclass · Answer

(B) Given the integral $I = \int \frac{1}{\left((x+4)^3(x+1)^5\right)^{1 / 4}} d x$. Rewrite the integrand as $I = \int \frac{1}{\left((x+4)^3(x+1)^3(x+1)^2\right)^{1 / 4}} d x = \int \frac{1}{\left((x+4)(x+1)\right)^{3/4} (x+1)^{2/4}} d x$. This simplifies to $I = \int \frac{1}{\left(\frac{x+4}{x+1}\right)^{3/4} (x+1)^2} d x$. Let $t = \frac{x+4}{x+1}$. Then $dt = \frac{(x+1)(1) - (x+4)(1)}{(x+1)^2} dx = \frac{-3}{(x+1)^2} dx$. Thus,$\frac{dx}{(x+1)^2} = -\frac{1}{3} dt$. Substituting these into the integral,we get $I = \int t^{-3/4} \left(-\frac{1}{3}\right) dt = -\frac{1}{3} \frac{t^{1/4}}{1/4} + c = -\frac{4}{3} \left(\frac{x+4}{x+1}\right)^{1/4} + c$. Comparing this with $A \cdot \left(\frac{x+4}{x+1}\right)^n + c$,we get $A = -\frac{4}{3}$ and $n = \frac{1}{4}$. Checking option $B$: $n + \frac{1}{A} = \frac{1}{4} + \frac{1}{-4/3} = \frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2}$. Thus,option $B$ is correct.

If $\int \frac{1}{\left((x+4)^3(x+1)^5\right)^{1 / 4}} d x=A \cdot\left(\frac{x+4}{x+1}\right)^n+c$,then which of the following is true?

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