Solve integral $\int \sqrt{3x^2 - 2x}\ dx$

Question

Find the following:

$$\int \sqrt{3x^2 - 2x}\ dx$$

I've tried completing the square and doing trigonometric substitution but I think I am making a mistake somewhere.

Thanks!

Answer 1

Hints;

Complete the square in the integrand to make it

$$\int \sqrt{\left(\sqrt3 x - \frac{1}{\sqrt3}\right )^2 -\frac13} \; \mathrm{d}x$$

substite $$u=\sqrt3 x - \frac{1}{\sqrt3}$$

Then substitute $$u=\frac{\sec(y)}{\sqrt3}$$

Don't forget that $$\sec^2(x)-1=\tan^2(x)$$ because you'll need to do it once from the LHS to RHS and then vice versa.

Finally, remember the reduction formula $$\int \sec^n(y) \; \mathrm{d}y = \frac{\sin(y) \sec^{n-1}(y)}{n-1}+\frac{n-2}{n-1}$$

And you're set. Enjoy :)