Question

Integrate the following functions.
$\int \frac{1}{9x^2+6x+5}dx.$

Answer 1

$\int \frac{1}{9x^2+6x+5}dx=\frac{1}{9}\int \frac{1}{x^2+\frac{2}{3}x+\frac{5}{9}}dx=\frac{1}{9}\int \frac{1}{x^2+2.\frac{1}{3}.x+(\frac{1}{3}{)}^2+\frac{5}{9}-(\frac{1}{3}{)}^2}dx=\frac{1}{9}\int \frac{1}{x+\frac{2}{3}x+(\frac{1}{3}{)}^2+\frac{5}{9}-\frac{1}{9}}dx=\frac{1}{9}\int \frac{1}{(x+\frac{1}{3}{)}^2+(\frac{2}{3}{)}^2}dx=\frac{1}{9}\left(\frac{1}{\frac{2}{3}}\right)ta{n}^{-1}\left(\frac{x+\frac{1}{3}}{\frac{2}{3}}\right)+C[\because \int \frac{dx}{a^2+x^2}=\frac{1}{a}ta{n}^{-1}(\frac{x}{a}\left)\right]=\frac{1}{6}ta{n}^{-1}\left(\frac{3x+1}{2}\right)+C$