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 

Your error stems from applying the chain rule incorrectly. You basically substitute $u=x\sqrt{2}\Rightarrow du=\sqrt{2}\cdot dx$.
So
$$\int\frac{dx}{(x\sqrt{2})^2+(\sqrt{5})^2}=\int\frac{\frac{du}{\sqrt{2}}}{u
^2+(\sqrt{5})^2}=\frac{1}{\sqrt{2}}\int\frac{du}{u
^2+(\sqrt{5})^2}=$$
$$=\frac{\arctan{\frac{u}{\sqrt{5}}}}{\sqrt{5}\mathbf{\sqrt{2}}}+C=\frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}\mathbf{\sqrt{2}}}+C$$
