int frac{x^3}{(x+1)^2} ,dx= | vedclass

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Question

$(A)$ Let $I = \int \frac{x^3}{(x+1)^2} dx$.
Substitute $u = x+1$,then $x = u-1$ and $dx = du$.
$I = \int \frac{(u-1)^3}{u^2} du = \int \frac{u^3 -

Vedclass · Accepted Answer

$\frac{x^2}{2}-2x+3\log|x+1|+\frac{1}{x+1}+c$

Vedclass · Answer

$(A)$ Let $I = \int \frac{x^3}{(x+1)^2} dx$.
Substitute $u = x+1$,then $x = u-1$ and $dx = du$.
$I = \int \frac{(u-1)^3}{u^2} du = \int \frac{u^3 - 3u^2 + 3u - 1}{u^2} du$.
$I = \int (u - 3 + \frac{3}{u} - \frac{1}{u^2}) du$.
Integrating term by term:
$I = \frac{u^2}{2} - 3u + 3\log|u| + \frac{1}{u} + c$.
Substituting $u = x+1$ back:
$I = \frac{(x+1)^2}{2} - 3(x+1) + 3\log|x+1| + \frac{1}{x+1} + c$.
$I = \frac{x^2+2x+1}{2} - 3x - 3 + 3\log|x+1| + \frac{1}{x+1} + c$.
$I = \frac{x^2}{2} + x + \frac{1}{2} - 3x - 3 + 3\log|x+1| + \frac{1}{x+1} + c$.
$I = \frac{x^2}{2} - 2x + 3\log|x+1| + \frac{1}{x+1} + C$ (where $C = c - 2.5$ is a constant).
Thus,the correct option is $A$.

$\int \frac{x^3}{(x+1)^2} \,dx=$

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