∫ x ^ 3 ( x+1 ) #160; ^ 2 #160; dx Substitute #160; x+1=t⇒ dt=dx =∫( t 1 ) #160; ^ 3 t ^ 2 #160; dt =∫ t ^ 3 1 3t ^ 2 +3t t ...

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Higher Order Derivatives

∫ x 3 / x+1 2...

Question

$\int \frac{x^{3}}{{(x + 1)}^{2}} d x$

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Solution

$\int \frac{x^{3}}{{(x + 1)}^{2}} d x$
Substitute $x + 1 = t \Rightarrow d t = d x$
$= \int \frac{{(t - 1)}^{3}}{t^{2}} d t$
$= \int \frac{t^{3} - 1 - {3 t}^{2} + 3 t}{t^{2}} d t$
$= \int [t - \frac{1}{t^{2}} - 3 + \frac{3}{t}] d t$
$= \int t d t - \int \frac{1}{t^{2}} d t - 3 \int d t + 3 \int \frac{1}{t} d t$
$= \frac{t^{2}}{2} + \frac{1}{t} - 3 t + 3 I n t + C$
$= \frac{{(x + 1)}^{2}}{2} + \frac{1}{x + 1} - 3 (x + 1) + 3 I n (x + 1) + C$

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