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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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    @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.2012-06-29

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

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    I was going to edit my post, but i guess you edited his for mine to be correct, thanks2012-06-29
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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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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!