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.
$$
\begin{align*}
\int \frac{dx}{16x^2+20x+35}
&=\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{35}{16}} \\
&=\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{10}{16}+\frac{25}{16}} \\
&=\frac{1}{16}\int \frac{dx}{(x+\frac{\sqrt{10}}{4})^2+(\frac{5}{4})^2}\\
&=\frac{1}{16}\frac{4}{5}\textstyle\arctan ((x+\frac{\sqrt{10}}{4})\cdot \frac{4}{5}) \\
&=\frac{1}{20}\textstyle\arctan(\frac{4x+\sqrt{10}}{5})
\end{align*}
$$
 A: 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?
A: 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!
A: Third equality is wrong. Try expanding it to see why it fails
