Question

Evaluate: $\int \frac{\frac{-1}{6}}{(x+3)}dx+\int \frac{\frac{1}{6}}{(x-3)}dx$

Answer 1

$I=\frac{-1}{6}\int \frac{dx}{x+3}+\frac{1}{6}\int \frac{dx}{x-3}$

$I=\frac{-1}{6}log(x+3)+\frac{1}{6}log(x-3)+c$

$I=\frac{1}{6}\left[log\right(x-3)-log(x+3\left)\right]+c$

$I=\frac{1}{6}log\left(\frac{x-3}{x+3}\right)+c$