Question

Solve : $\int \frac{1}{9x^2-1}dx$

Answer 1

$\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+c$

$\int \frac{1}{9x^2-1}dx$

$=\frac{1}{9}\int \frac{1}{x^2-\frac{1}{9}}dx$

$=\frac{1}{9}\int \frac{1}{x^2-(\frac{1}{3}{)}^2}dx$

$=\frac{1}{9}\times \frac{3}{2}\log \left|\frac{x-\frac{1}{3}}{x+\frac{1}{3}}\right|+c$

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