Evaluate the following integrals: 1-3x 3x^2+4x+2 dx

Question

Evaluate the following integrals:

$\int\frac{1-3\text{x}}{3\text{x}^2+4\text{x}+2}\text{ dx}$

✓

Answer

Let $\text{I}=\int\frac{1-3\text{x}}{3\text{x}^2+4\text{x}+2}\text{ dx}$
Let $1-3\text{x}=\lambda\frac{\text{d}}{\text{dx}}\big(3\text{x}^2+4\text{x}+2\big)+\mu$
$=\lambda(6\text{x}+4)+\mu$
$1-3\text{x}=(6\lambda)\text{x}+(4\lambda+\mu)$
Comparing the coefficients of like powers of x,
$6\lambda=3\Rightarrow\lambda=-\frac{1}{2}$
$4\lambda+\mu=1\Rightarrow4\Big(-\frac{1}{2}\Big)+\mu=1$
$\mu=3$
So, $\text{I}=\int\frac{-\frac{1}{2}(6\text{x}+4)+3}{3\text{x}^2+4\text{x}+2}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+3\int\frac{1}{3\text{x}^2+4\text{x}+2}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+\frac{3}{3}\int\frac{1}{\text{x}^2+\frac{4}{3}\text{x}+\frac{2}{3}}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{2}{3}\big)+\big(\frac{2}{3}\big)^2-\big(\frac{2}{3}\big)^2+\frac{2}{3}}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+\int\frac{1}{\big(\text{x}+\frac{2}{3}\big)^2+\frac{2}{9}}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+\int\frac{1}{\big(\text{x}+\frac{2}{3}\big)^2+\big(\frac{\sqrt2}{3}\big)^2}\text{ dx}$
$\text{I}=-\frac{1}{2}\log\big|3\text{x}^2+4\text{x}+2\big|+\frac{3}{\sqrt2}\tan^{-1}\bigg(\frac{\text{x}+\frac{2}{3}}{\frac{\sqrt2}{3}}\bigg)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\frac{1}{2}\log\big|3\text{x}^2+4\text{x}+2\big|+\frac{3}{\sqrt2}\tan^{-1}\Big(\frac{3\text{x}+2}{\sqrt2}\Big)+\text{C}$