If int frac{x}{(a+x)^5} dx = frac{1}{k(a+x)^4 | vedclass

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(D) Let $u = x+a$, then $du = dx$ and $x = u-a$. Substituting these into the integral: $\int \frac{u-a}{u^5} du = \int (u^{-4} - au^{-5}) du$ $= \frac

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$\frac{1}{4}$

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(D) Let $u = x+a$, then $du = dx$ and $x = u-a$. Substituting these into the integral: $\int \frac{u-a}{u^5} du = \int (u^{-4} - au^{-5}) du$ $= \frac{u^{-3}}{-3} - a \frac{u^{-4}}{-4} + C$ $= -\frac{1}{3u^3} + \frac{a}{4u^4} + C$ $= \frac{-4u + 3a}{12u^4} + C$ $= \frac{-4(x+a) + 3a}{12(x+a)^4} + C$ $= \frac{-4x - 4a + 3a}{12(x+a)^4} + C$ $= \frac{1}{12(x+a)^4}(-4x - a) + C$ Comparing this with $\frac{1}{k(a+x)^4}(f(x)) + c$, we get $k = 12$ and $f(x) = -4x - a$. Now, calculate $f(-a) = -4(-a) - a = 4a - a = 3a$. Finally, $\frac{f(-a)}{ak} = \frac{3a}{12a} = \frac{3}{12} = \frac{1}{4}$.

If $\int \frac{x}{(a+x)^5} dx = \frac{1}{k(a+x)^4}(f(x)) + c$, then $\frac{f(-a)}{ak} = $

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