Problem with computing $\int\frac{dx}{2x^2+5} $ by trigonometric substitution

Question

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

Answer 1

Your error stems from applying the chain rule incorrectly. You basically substitute $u=x\sqrt{2}\Rightarrow du=\sqrt{2}\cdot dx$.

So

$$\int\frac{dx}{(x\sqrt{2})^2+(\sqrt{5})^2}=\int\frac{\frac{du}{\sqrt{2}}}{u ^2+(\sqrt{5})^2}=\frac{1}{\sqrt{2}}\int\frac{du}{u ^2+(\sqrt{5})^2}=$$

$$=\frac{\arctan{\frac{u}{\sqrt{5}}}}{\sqrt{5}\mathbf{\sqrt{2}}}+C=\frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}\mathbf{\sqrt{2}}}+C$$

Answer 2

Setting $x\mapsto\ \sqrt{\frac{5}{2}}u$ yields: $$\int_{ }^{ }\frac{dx}{2x^{2}+5}=\sqrt{\frac{1}{10}}\int_{ }^{ }\frac{1}{u^{2}+1}du=\sqrt{\frac{1}{10}}\arctan\left(u\right)+C=\sqrt{\frac{1}{10}}\arctan\left(\sqrt{\frac{2}{5}}x\right)+C$$