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Possible Duplicate:
Help evaluating $\int \frac{dx}{(x^2 + a^2)^2}$

How to I begin this integration problem?

$\begin{align}\int_{0}^{1} \frac{dx}{{\left(x^2 + 1\right)}^{2}}\end{align}$

I'm not really sure how setup the triangle to do trigonometric substitution for this problem, since I am squaring the bottom, not taking the square root of it.

Could someone please demonstrate/explain how I could do this?

Thank you for your time.

  • 0
    Whoops, fixed. ;)2012-03-12

3 Answers 3

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There is a "trick" here: begin noticing that $1=x^2+1-x^2$ so $\frac 1{(x^2+1)^2}=\frac{x^2+1-x^2}{(x^2+1)^2}=\frac 1{x^2+1}-\frac {x^2}{(x^2+1)^2}.$ A primitive of the first term is well-known, for the second we integrate by parts: $\int\frac{x\cdot x}{(x^2+1)^2}dx=-\frac x2\frac 1{1+x^2}+\frac 12\int \frac 1{1+x^2}dx$ so finally $\int \frac{dx}{(1+x^2)^2}=\frac 12\arctan x+\frac x2\frac 1{1+x^2}.$

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The standard substitution with $x^2 + 1$ in the denominator is $x = \tan \theta$.

With this we get

$dx = \sec^2 \theta d\theta$

and so your integral is

$\int_{0}^{\pi/4} \frac{1}{1 + \tan^2 \theta} d\theta = \int_{0}^{\pi/4} \cos^2 \theta d \theta$

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    The point is that the square root is not essential to the use of trigonometric substitutions. They are worth trying on integrals containing expressions of the form $x^2 \pm a^2$ or $a^2 - x^2$ , $a^2$ being some non-zero constant. (Sometimes simpler techniques will do, of course.)2013-04-21
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Hint: Let $x=\tan(y)$, then $dx=\sec^2(y) dy$, and $x^2+1$ will be then $\tan^2(y)+1=sec^2y$. Can you finish it from here?