How do you integrate $\frac{x^{3}}{{(x^{2} + 5)}^{2}}$ $x^3/((x^2+5)^2)$ ?

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How do you integrate $\frac{x^{3}}{{(x^{2} + 5)}^{2}}$ $x^3/((x^2+5)^2)$ ?

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Answer 1 · 2015-03-26T22:19:04

$\int \frac{x^{3}}{{(x^{2} + 5)}^{2}} d x$ $int x^3/(x^2+5)^2dx$

We can rewrite :

$\frac{x^{3}}{{(x^{2} + 5)}^{2}} = \frac{β x + γ}{x^{2} + 5} + \frac{θ x + ν}{{(x^{2} + 5)}^{2}}$ $x^3/(x^2+5)^2= (betax+gamma)/(x^2+5) + (thetax+nu)/(x^2+5)^2$

So : $\frac{(β x + γ) (x^{2} + 5)}{(x^{2} + 5) (x^{2} + 5)} + \frac{θ x + ν}{{(x^{2} + 5)}^{2}}$ $((betax+gamma)(x^2+5))/((x^2+5)(x^2+5))+(thetax+nu)/(x^2+5)^2$

Then : $\frac{β x^{3} + γ x^{2} + (5 β + θ) x + ν + 5 γ}{{(x^{2} + 5)}^{2}}$ $(betax^3+gammax^2+(5beta+theta)x+nu+5gamma)/(x^2+5)^2$

Identification :

$ν + 5 γ = 0$ $nu + 5gamma = 0$
$5 β + θ = 0$ $5beta + theta = 0$
$γ = 0$ $gamma = 0$
$β = 1$ $beta = 1$

So :

$β = 1$ $beta = 1$
$γ = 0$ $gamma = 0$
$θ = - 5$ $theta = -5$
$ν = 0$ $nu = 0$

$\int \frac{x^{3}}{{(x^{2} + 5)}^{2}} d x = \int \frac{x}{x^{2} + 5} d x - 5 \int \frac{x}{{(x^{2} + 5)}^{2}} d x$ $int x^3/(x^2+5)^2dx = intx/(x^2+5)dx-5int(x)/(x^2+5)^2dx$

$\frac{1}{2} \int \frac{2 x}{x^{2} + 5} d x - \frac{5}{2} \int \frac{2 x}{{(x^{2} + 5)}^{2}} d x$ $1/2int (2x)/(x^2+5)dx -5/2int(2x)/(x^2+5)^2dx$

$= (\frac{1}{2} ln (x^{2} + 5) + \frac{5}{2 x^{2} + 10}) + C$ $= (1/2ln(x^2+5)+5/(2x^2+10))+C$

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How do you integrate $\frac{x^{3}}{{(x^{2} + 5)}^{2}}$ $x^3/((x^2+5)^2)$ ?

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