How do you evaluate the integral $\int \frac{d x}{{(x + 1)}^{3}}$ $int dx/(x+1)^3$ ?

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Answer 1 · 2017-01-15T20:37:59

$\int \frac{d x}{{(x + 1)}^{3}} = \frac{- 1}{2 {(x + 1)}^{2}} + C$ $intdx/(x+1)^3=(-1)/(2(x+1)^2)+C$

Let $u = x + 1$ $u=x+1$ . Differentiating this shows that $d u = d x$ $du=dx$ . We can then write:

$\int \frac{d x}{{(x + 1)}^{3}} = \int \frac{d u}{u^{3}} = \int u^{- 3} d u$ $intdx/(x+1)^3=int(du)/u^3=intu^-3du$

This can be integrated through the rule $\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C$ $intu^ndu=u^(n+1)/(n+1)+C$ where $n \neq - 1$ $n!=-1$ .

$\int \frac{d x}{{(x + 1)}^{3}} = \frac{u^{- 2}}{- 2} + C = \frac{- 1}{2 u^{2}} + C$ $intdx/(x+1)^3=u^(-2)/(-2)+C=(-1)/(2u^2)+C$

Returning to the original variable $x$ $x$ with $u = x + 1$ $u=x+1$ :

$\int \frac{d x}{{(x + 1)}^{3}} = \frac{- 1}{2 {(x + 1)}^{2}} + C$ $intdx/(x+1)^3=(-1)/(2(x+1)^2)+C$

How do you evaluate the integral ∫dx(x+1)3int dx/(x+1)^3?