<s>[INST] Problem with computing $\int\frac{dx}{2x^2+5} $ by trigonometric substitution I so close to can solve this problem but I don't find the correct response:
$$\int\frac{dx}{2x^2+5}
$$
Always get the answer:
$$ \frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}}
$$
But the correct answer have more one square root multiplying the square root of $5:$
(this is the correct answer):
$$
\frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}\mathbf{\sqrt{2}}}
$$
I'm using the follow propertie to try solve the problem:
$$
\int{\frac{dx}{u^2+a^2}} = \frac{1}{a}\arctan{\frac{u}{a}}
$$
--My steps for the solution:
Before to apply the properties get square root from $~2x^2~$ and $5$, staying that way:
$$\int\frac{dx}{(x\sqrt{2})^2+(\sqrt{5})^2}
$$
So when I apply the previous properties get my wrong answer 
 [/INST] Setting $x\mapsto\ \sqrt{\frac{5}{2}}u$ yields:
$$\int_{ }^{ }\frac{dx}{2x^{2}+5}=\sqrt{\frac{1}{10}}\int_{ }^{ }\frac{1}{u^{2}+1}du=\sqrt{\frac{1}{10}}\arctan\left(u\right)+C=\sqrt{\frac{1}{10}}\arctan\left(\sqrt{\frac{2}{5}}x\right)+C$$
</s>