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$$\int{\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}} dx$$

So ... how do I start? Numerator cant be factorized it seems, and this looks like a complicated expression ...

I tried expanding the denominator to see if integration by substitution will work, but it didn't give any ideas

$$\int{\frac{5x^3+8x^2+x+2}{2x^4+x^2}} dx$$

I don't see a clear way to integrate by parts either. Can anyone enlighten me?

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Partial fractions. –  Hans Lundmark Apr 8 '12 at 13:42
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3 Answers

up vote 5 down vote accepted

Hint :

Use partial fraction decomposition :

$$\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{2x^2+1}$$

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If you wish to avoid partial fractions, it is possible to clean up that denominator with a simple substitution. To make that $2x^2+1$ factor more manageable, make the substitution

$x=\frac{\sqrt2}2\tan\theta,dx=\frac{\sqrt2}2\sec^2\theta d\theta$

$\int\dfrac{5x^3+8x^2+x+2}{x^2(2x^2+1)}dx=\int\dfrac{\frac{5\sqrt2}4\tan^3\theta+4\tan^2\theta+\frac{\sqrt2}2\tan\theta+2}{\frac12\tan^2\theta(\tan^2\theta+1)}\times\frac{\sqrt2}2\sec^2\theta d\theta$

$\dfrac{\frac{\sqrt2}2\sec^2\theta}{\frac12(\tan^2+1)}$ reduces to $\sqrt2$. Then dividing through gets

$\int\frac52\tan\theta+4\sqrt2+\cot\theta+2\sqrt2\cot^2\theta d\theta$

Can you take it from there?

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You can make descomposition of the quotients as a sum of simple fractions. $$\dfrac{A}{x} + \dfrac {B}{x^{2}} + \dfrac{Cx+D}{2x^2+1}.$$

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