Question

$\int \frac{1}{3x^2+13x-10}dx$

Answer 1

$$\begin{gathered} \int \frac{1}{3x^2+13x-10}dx \\ =\frac{1}{3}\int \frac{1}{x^2+{\displaystyle \frac{13}{3}}x-{\displaystyle \frac{10}{3}}}dx \\ =\frac{1}{3}\int \frac{1}{x^2+{\displaystyle \frac{13x}{3}}+{\left({\displaystyle \frac{13}{6}}\right)}^2-{\left({\displaystyle \frac{13}{6}}\right)}^2-{\displaystyle \frac{10}{3}}}dx \\ =\frac{1}{3}\int \frac{1}{{\left(x+{\displaystyle \frac{13}{6}}\right)}^2-{\displaystyle \frac{169}{36}}-{\displaystyle \frac{10}{3}}}dx \\ =\frac{1}{3}\int \frac{1}{{\left(x+{\displaystyle \frac{13}{6}}\right)}^2-{\displaystyle \frac{169-120}{36}}}dx \\ =\frac{1}{3}\int \frac{1}{{\left(x+{\displaystyle \frac{13}{6}}\right)}^2-{\left({\displaystyle \frac{17}{6}}\right)}^2}dx \\ =\frac{1}{3}\times \frac{1}{2\times {\displaystyle \frac{17}{6}}}\ln \left|\frac{x+{\displaystyle \frac{13}{6}}-{\displaystyle \frac{17}{6}}}{x+{\displaystyle \frac{13}{6}}+{\displaystyle \frac{17}{6}}}\right|\left[\because \int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+C\right] \\ =\frac{1}{17}\ln \left|\frac{x-{\displaystyle \frac{2}{3}}}{x+5}\right|+C \\ =\frac{1}{17}\ln \left|\frac{3x-2}{3x+15}\right|+C \end{gathered}$$