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I've attempted to integrate the function $\frac{x^2+1}{(x^2-2x+2)^2}$ using several techniques, but none of them are solving it nicely.

I know there must be a way to do this using trig substitution (its from that chapter of my text book), just not sure how to do it..

I've tried substituting and simplifying till:

$$\frac{(1+u)^2+1}{(1+u^2)^2}$$

Then subbing $1+u^2$ for $\sec^2(u)$ but still no luck. Anything I'm missing?

I think it might be possible to rewrite the initial function to have a $\sqrt{\cdot}$ as the denominator but not sure how to work that out...

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2 Answers

up vote 4 down vote accepted

Here is the trigonometric approach

$$ \int \frac{x^2+1}{(x^2-2x+2)^2}dx = \int \frac{x^2+1}{((x-1)^2+1)^2}dx = \int\frac{(1+u)^2+1}{(u^2+1)^2}du .$$

Using the substitution $u=\tan(y)$ gives

$$ \int \!{\frac { \tan\left( y \right)^{2}+2\,\tan \left( y \right) +2}{ (1+\tan\left(y\right)^{2})^2 }}{dy}$$

$$ = 2\int(\sin \left( y \right) \cos \left( y \right) + \cos \left( y \right)^{2}+1)\,{dy} =\dots $$

Notes:

$$ (1)\,\, 1+\tan(y)^2=\sec(y)^2, $$

$$ (2)\,\, \cos(2y) = 2\cos(y)^2 - 1. $$

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You lost an exponent in the denominator. –  Mike Feb 9 '13 at 6:51
    
@Mike: It is done. Thank you for the comment. –  Mhenni Benghorbal Feb 9 '13 at 16:09
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As a first step one can use partial fractions to express our integrand as $$\frac{A+Bu}{1+u^2}+\frac{C+Du}{(1+u^2)^2}.$$

Then the $Bu$ and $Cu$ yield to a simple substitution, and for the $A$ part we will get an arctan. It remains to integrate $\frac{C}{(1+u^2)^2}$. This can be done by a trigonometric substitution, or by using integration by parts.

The trigonometric substitution is the obvious $u=\tan\theta$. We then end up integrating $\cos^2\theta$. I somewhat prefer parts. The idea is to attempt to integrate $\frac{1}{1+u^2}$ (yes) by parts, letting $dv=du$ and $w=\frac{1}{1+u^2}$.

Another way: The first solution was generic, and intended to show that there is a general algorithmic procedure for such problems.

We can do it with much less machinery. The numerator is $1+u^2 +2u+1$. When we divide by $(1+u^2)^2$, we get $$\frac{1}{1+u^2}+\frac{2u}{1+u^2} +\frac{1}{(1+u^2)^2}.$$ Integrate. The first term gives $\arctan u$. The second, possibly after letting $1+u^2=v$, gives $\ln(1+u^2)$.

It remains to deal with the third term. Let $u=\tan t$. After the smoke clears, we want $\int \cos^2 t\,dt$, which is a standard integral.

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Thanks for the help, this was the only way I had found the right answer. Problem is, integration of partial fractions comes after trig sub in the text book, so there has to be another way still :p –  RyanS Feb 9 '13 at 3:51
    
The first part of the answer above was the "generic" answer. For your particular problem, I have sketched a simpler way. If it is not explicit enough to follow, I will write the whole thing out. –  André Nicolas Feb 9 '13 at 4:03
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