Problem asked by Yulyasy Ariyanto

Evaluate $\int \frac{2x-1}{\sqrt{x^2+6x + 5}} \; dx $

Answer:

$\int \frac{2x-1}{\sqrt{x^2+6x + 5}} \; dx $

$=\int \frac{2x-1}{\sqrt{(x+3)^2 -4}} \; dx $

Substitute $u = x + 3$ and $du = dx$, and we have

$\int \frac{2u-7}{\sqrt{u^2 -4}} \; du $

Substitute $u = 2\sec \theta$ and $du = 2\sec \theta \tan \theta \; d\theta $, and we have

$\int \frac{4\sec \theta-7}{2\tan \theta} \; 2\sec \theta \tan \theta \; d\theta $

$\int (4\sec \theta-7) \sec \theta \; d\theta $

$\int (4\sec^2 \theta - 7\sec \theta) \; d\theta $

$=4\tan \theta - 7\ln |\sec \theta + \tan \theta| +C $

$=2\sqrt{x^2 + 6x + 5} - 7\ln \left| \frac{x+3}{2} + \frac{\sqrt{x^2 + 6x + 5}}{2} \right| + C $

$=2\sqrt{x^2 + 6x + 5} - 7\ln \left| x+3+ \sqrt{x^2 + 6x + 5} \right| + C $