Question

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

Answer 1

$$\begin{gathered} \int \left(\frac{x-1}{3x^2-4x+3}\right)dx \\ x-1=\mathrm{A}\frac{d}{dx}\left(3x^2-4x+3\right)+\mathrm{B} \\ x-1=\mathrm{A}\left(6x-4\right)+\mathrm{B} \\ x-1=\left(6\mathrm{A}\right)x+\mathrm{B}-4\mathrm{A} \end{gathered}$$

Comparing the Coefficients of like powers of x
$$\begin{gathered} 6\mathrm{A}=1 \\ \mathrm{A}=\frac{1}{6} \\ \mathrm{B}-4\mathrm{A}=-1 \\ \mathrm{B}-4\times \frac{1}{6}=-1 \\ \mathrm{B}=-1+\frac{2}{3} \\ \mathrm{B}=\frac{1}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\int \frac{\left(x-1\right)dx}{3x^2-4x+3} \\ =\int \left[\frac{{\displaystyle \frac{1}{6}}\left(6x-4\right)+{\displaystyle \frac{1}{3}}}{3x^2-4x+3}\right]dx \\ =\frac{1}{6}\int \frac{\left(6x-4\right)dx}{3x^2-4x+3}+\frac{1}{3}\int \frac{dx}{3x^2-4x+3} \\ =\frac{1}{6}\int \frac{\left(6x-4\right)dx}{3x^2-4x+3}+\frac{1}{9}\int \frac{dx}{x^2-{\displaystyle \frac{4}{3}}x+1} \\ =\frac{1}{6}\int \frac{\left(6x-4\right)dx}{3x^2-4x+3}+\frac{1}{9}\int \frac{dx}{x^2-{\displaystyle \frac{4}{3}}x+{\left({\displaystyle \frac{2}{3}}\right)}^2{\left({\displaystyle \frac{2}{3}}\right)}^2+1} \\ =\frac{1}{6}\int \frac{\left(6x-4\right)dx}{3x^2-4x+3}+\frac{1}{9}\int \frac{dx}{{\left(x-{\displaystyle \frac{2}{3}}\right)}^2-{\displaystyle \frac{4}{9}}+1} \\ =\frac{1}{6}\int \frac{\left(6x-4\right)dx}{3x^2-4x+13}+\frac{1}{9}\int \frac{dx}{{\left(x-{\displaystyle \frac{2}{3}}\right)}^2+{\left({\displaystyle \frac{\sqrt{5}}{3}}\right)}^2} \\ =\frac{1}{6}\log \left|3x^2-4x+3\right|+\frac{1}{9}\times \frac{3}{\sqrt{5}}{\tan }^{-1}\left(\frac{x^{-{\displaystyle \frac{2}{3}}}}{{\displaystyle \frac{\sqrt{5}}{3}}}\right)+C \\ =\frac{1}{6}\log \left|3x^2-4x+3\right|+\frac{1}{3\sqrt{5}}{\tan }^{-1}\left(\frac{3x-2}{\sqrt{5}}\right)+C \\ =\frac{1}{6}\log \left|3x^2-4x+3\right|+\frac{\sqrt{5}}{15}{\tan }^{-1}\left(\frac{3x-2}{\sqrt{5}}\right)+C \end{gathered}$$