Question:
$\sqrt{x^{2}+4 x-5}$
Solution:
Let $I=\int \sqrt{x^{2}+4 x-5} d x$
$=\int \sqrt{\left(x^{2}+4 x+4\right)-9} d x$
$=\int \sqrt{(x+2)^{2}-(3)^{2}} d x$
It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+\mathrm{C}$
$\therefore I=\frac{(x+2)}{2} \sqrt{x^{2}+4 x-5}-\frac{9}{2} \log \left|(x+2)+\sqrt{x^{2}+4 x-5}\right|+\mathrm{C}$
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