# Integration: How to Begin? [duplicate]

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.

-

## marked as duplicate by Sivaram Ambikasaran, Davide Giraudo, Fabian, Alex Becker, mixedmath♦Mar 12 '12 at 23:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What a coincidence that this question was posted on the same day as mine. Wonder if we're working from the book, at the same time, (from the same school, in the same class)? lol Thank you, Sivaram – Oliver Spryn Mar 12 '12 at 22:33
@SivaramAmbikasaran: Since this is a definite integral, we might potentially have different solutions, so I am not so sure we should close this as a dupe. – Aryabhata Mar 12 '12 at 22:38
@Aryabhata Once we have the indefinite integral, it is just a question of plugging in the limits and hence I voted to close it. The OP too seems to acknowledge this. – user17762 Mar 12 '12 at 23:07
@SivaramAmbikasaran: Yes, but that does not mean there aren't other solutions. For instance if the limits were $0$ to $\infty$, a substitution of $x = \frac{1}{t}$ and adding gives us the answer easily. – Aryabhata Mar 12 '12 at 23:09
@Aryabhata True. I get it. I concede. Is there any way I can undo? – user17762 Mar 12 '12 at 23:29

## 3 Answers

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 $$\int \frac{dx}{(1+x^2)^2}=\frac 12\arctan x+\frac x2\frac 1{1+x^2}.$$

-

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

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

-
The only problem is, that I don't understand why this works (or really how to set it up) since we aren't working with a square root here. – Oliver Spryn Mar 12 '12 at 23:36
@spryno724: Why do you want a square root? I think you should stop thinking in terms of right triangles! This is just integration by substitution. For every trigonometric substitution, if you start drawing a right triangle, this is going to be really inefficient. – Aryabhata Mar 12 '12 at 23:37
Hmm... how can we stop thinking in terms of right triangles if the above example uses $\tan{x}$ and $\cos{x}$? Plus, by having a square root, we easily know that it would (in this case) exist as the hypotenuse of the triangle. – Oliver Spryn Mar 12 '12 at 23:41
@spryno724: The identity used here is $1 + \tan^2 \theta = \sec^2 \theta$, which is basically a rewritten version of $\sin^2 \theta + \cos^2 \theta = 1$. I guess a little more practice will make this easier. If thinking in terms of right triangles is easier for you, go ahead. But do try to avoid square roots if not required :-) – Aryabhata Mar 12 '12 at 23:49
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.) – RecklessReckoner Apr 21 '13 at 22:45

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?

-