Question

Solve $\int \frac{1}{9x^2+6x+10}dx$

Answer 1

$\int \frac{1}{9x^2+6x+10}dx=\int \frac{1}{9x^2+6x+1+9}dx$

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

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

$=\frac{1}{9}\int \frac{1}{{1+(x+\frac{1}{3})}^2}d(x+\frac{1}{3})$

$=\frac{1}{9}{\tan }^{-1}(x+\frac{1}{3})+C$