Answer
$$\frac{1}{5} \ln \left|\frac{5 x+\sqrt{25 x^{2}+2}}{\sqrt{2}}\right|+C$$
Work Step by Step
Given $$\int \frac{d x}{\sqrt{25 x^{2}+2}} $$
Let $$x=\frac{\sqrt{2}}{5} \tan \theta, \quad d x=\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta $$
Then
\begin{aligned}
\int \frac{d x}{\sqrt{25 x^{2}+2}} &=\int \frac{\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta}{\sqrt{25\left(\frac{\sqrt{2}}{5} \tan \theta\right)^{2}+2}} \\
&=\int \frac{\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta}{\sqrt{25 \cdot \frac{2}{28} \tan ^{2} \theta+2}} \\
&=\int \frac{\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta}{\sqrt{2\left(\tan ^{2} \theta+1\right)}} \\
&=\int \frac{\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta}{\sqrt{2} \sqrt{\sec ^{2} \theta}}\\
&=\frac{1}{5} \int \sec \theta d \theta\\
&=\frac{1}{5} \ln |\sec \theta+\tan \theta|+C\\
&=\frac{1}{5} \ln \left|\frac{5 x+\sqrt{25 x^{2}+2}}{\sqrt{2}}\right|+C
\end{aligned}
(Note that the $\sqrt 2$ can be combined into the constant C as well.)
