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

$$\rightarrow \int \left(x+ \frac{2}{x+1} - \frac{1}{\left(x+1\right)^2} -2 \right)\, \mathrm{d}x$$

is there another method other than long division to solve this integral?

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The $-x$ and the $-1$ of the numerator are handled by $u$-subsitution and a tangent substitution. Probably the other can be similarly dispatched. Of course, the resulting trig. integrals may well be worse than the solution route you indicate in your post. – James S. Cook Apr 8 '13 at 0:54
What happened to the $x^2$ in the denominator? As written, the numerator has lesser degree than the denominator so division isn't likely to yield anything. – Mike Apr 8 '13 at 1:32
@Mike, thanks, fixed the typo. – yiyi Apr 8 '13 at 4:42
up vote 4 down vote accepted

You can do a trig substitution: let $x=\tan{\theta}$, $dx = \sec^2{\theta}\,d\theta$. You get

$$\int d\theta \: [\tan{\theta}(1-2 \cos^2{\theta}) - \cos^2{\theta}]$$

which comes out to

$$-\log{\cos{\theta}} + \frac{1}{2} \cos{2 \theta} - \frac{\theta}{2} - \frac{1}{4} \sin{2 \theta} + C$$

where $C$ is an integration constant. You then substitute back, details of which I leave to the reader.

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Why did you write the equation behind the $\mathrm{d}\theta$? – yiyi Apr 8 '13 at 1:15
It's the way I got used to writing integrals. It comes from my background in Optics, where we have multiple integrals commonly and it is useful to write integrals like that to keep track of which integral goes with which variable. – Ron Gordon Apr 8 '13 at 1:16

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