# Find the indefinite integral of $1/(16x^2+20x+35)$

Here is my steps of finding the integral, the result is wrong but I don't know where I made a mistake or I may used wrong method.

\begin{align*} \int \frac{dx}{16x^2+20x+35} &=\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{35}{16}} \\ &=\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{10}{16}+\frac{25}{16}} \\ &=\frac{1}{16}\int \frac{dx}{(x+\frac{\sqrt{10}}{4})^2+(\frac{5}{4})^2}\\ &=\frac{1}{16}\frac{4}{5}\textstyle\arctan ((x+\frac{\sqrt{10}}{4})\cdot \frac{4}{5}) \\ &=\frac{1}{20}\textstyle\arctan(\frac{4x+\sqrt{10}}{5}) \end{align*}

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I'll try to clean up your post. In the meanwhile, have you read the sptes? wolframalpha.com/input/?i=Integrate[1%2F%2816+x^2%2B20+x%2B35%29%2Cx] –  Siminore Jun 29 '12 at 14:01
Wow..This is the first time the "show step" bottom gives me some steps instead of putting an equal sign between the question and result. I read the steps on wolfram alpha but I am trying to use the method from my textbook, the book said "Always reduce the coefficient of x^2 to unity before integrating". btw, thank you for cleaning up the post, could you give me a link or something about how to make the post in better format? –  Vic. Jun 29 '12 at 14:13
@Vic: re: formatting, search for "LaTeX tutorial", but a pretty good way to learn is to just look at what other people have done: you can right-click on any formula and get a little menu, select Show Math As → TeX commands, and it'll show you what you have to write (in between a pair of dollar signs, or double dollar signs) to get that result. –  Ben Millwood Jun 29 '12 at 14:20

Your problem is this step: $$\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{10}{16}+\frac{25}{16}} =\frac{1}{16}\int \frac{dx}{(x+\frac{\sqrt{10}}{4})^2+(\frac{5}{4})^2}$$ for which you use this equality: $$\textstyle x^2+\frac{20}{16}x+\frac{10}{16}+\frac{25}{16} =(x+\frac{\sqrt{10}}{4})^2+(\frac{5}{4})^2$$ but that's just not true! The right-hand side expands into $x^2 + 2\frac{\sqrt{10}}{4}x + \frac{10}{16} + \frac{25}{16}$: as you can see the $x$ term is wrong, and the square root is unnecessary. Just to remind you, the general rule for completing the square is: $$\textstyle x^2 + bx + c = (x + \frac{b}{2})^2 + c - \frac{b^2}{4}$$

No square roots anywhere!

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You want to complete a square. So, remember that $$(x+\alpha)^2 = x^2 + 2 \alpha x + \alpha^2.$$ You have $$2\alpha = \frac{20}{16},$$ i.e. $\alpha = 5/8$. Hence $$x^2 +\frac{20}{16}x = \left( x + \frac{5}{8} \right)^2 - \frac{25}{64}.$$ Can you go further, now?

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Third equality is wrong. Try expanding it to see why it fails

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Notice that the post has been edited a whole bunch of times, so you might want to be sure you're addressing his mistake and not one that was made later. –  Ben Millwood Jun 29 '12 at 14:14
I was going to edit my post, but i guess you edited his for mine to be correct, thanks –  Jean-Sébastien Jun 29 '12 at 14:23