# How do you prove the integral formula intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C ?

Sep 11, 2014

Let $x = a \tan \theta$. $R i g h t a r r o w \mathrm{dx} = a {\sec}^{2} \theta d \theta$
So, we can write
$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} + {a}^{2}}} = \int \frac{a {\sec}^{2} \theta}{\sqrt{{a}^{2} \left({\tan}^{2} \theta + 1\right)}} d \theta$
by ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$,
$= \int \sec \theta d \theta = \ln | \sec \theta + \tan \theta | + C$
since $\tan \theta = \frac{x}{a}$ and $\sec \theta = \frac{\sqrt{{x}^{2} + {a}^{2}}}{a}$,
$= \ln | \frac{\sqrt{{x}^{2} + {a}^{2}} + x}{a} | + {C}_{1}$
by the log property $\ln \left\{\frac{A}{B}\right\} = \ln A - \ln B$,
$= \ln | \sqrt{{x}^{2} + {a}^{2}} + x | - \ln | a | + {C}_{1}$
by setting $C = \ln | a | + {C}_{1}$,
$= \ln | x + \sqrt{{x}^{2} + {a}^{2}} | + C$