Antiderivative of root as numerator.

Question

Just learned about substitution. However, I find it confusing when the root is at the numerator of the fraction and gets complicated when I substituted it.

$$\int {\sqrt{2x^2-4}\over 7x} dx$$

Is there any way to get around it or other method to use?

Answer 1

Put $\tan u=\sqrt{\frac{x^2}{2}-1}$.

Note that $x^2=2\sec^2u$ and $2xdx=4\sec^2 u\tan udu$.

$$ \begin{align*} \int\frac{\sqrt{2x^2-4}}{7x}dx &=\frac{1}{7}\int\frac{\sqrt{\frac{x^2}{2}-1}}{x^2}\cdot 2xdx \\ &=\frac{1}{7}\int\frac{\tan u}{2\sec^2u}\cdot 4\sec^2 u\tan udu \\ &=\frac{2}{7}\int\tan^2 udu \\ &=\frac{2}{7}\int(\sec^2 u-1)du \\ &=\frac{2}{7}(\tan u-u)+C \\ &=\frac{2}{7}\left[\sqrt{\frac{x^2}{2}-1}-\tan^{-1}\left(\sqrt{\frac{x^2}{2}-1}\right)\right]+C \\ \end{align*} $$

Answer 2

I try to tackle the integral by “rationalisation” followed by integration by parts as below: $$ \begin{aligned} \int \frac{\sqrt{2 x^2-4}}{7 x} d x&=\frac{1}{14} \int \frac{2 x^2-4}{x^2} d \left(\sqrt{2 x^2-4} \right)\\ & =\frac{1}{7} \int d \left(\sqrt{2 x^2-4}\right)-\frac{2}{7} \int \frac{d \left(\sqrt{2 x^2-4}\right)}{x^2} \\ & =\frac{1}{7} \sqrt{2 x^2-4}-\frac{4}{7} \int \frac{d\left(\sqrt{2 x^2-4}\right)}{\left(\sqrt{2 x^2-4}\right)^2+4} \\ & =\frac{1}{7} \sqrt{2 x^2-4}-\frac{2}{7} \arctan \left(\frac{\sqrt{2 x^2-4}}{2}\right)+C \end{aligned} $$