Question

$\int \frac{1}{x\left(x-2\right)\left(x-4\right)}dx$

Answer 1

$$\begin{gathered} \int \frac{1}{x\left(x-2\right)\left(x-4\right)}dx \\ \mathrm{Let}\frac{1}{x\left(x-2\right)\left(x-4\right)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-4} \\ \Rightarrow \frac{1}{x\left(x-2\right)\left(x-4\right)}=\frac{A\left(x-2\right)\left(x-4\right)+B\left(x\right)\left(x-4\right)+Cx·\left(x-2\right)}{x\left(x-2\right)\left(x-4\right)} \\ \Rightarrow 1=A\left(x-2\right)\left(x-4\right)+B\left(x\right)·\left(x-4\right)+Cx.\left(x-2\right)...\left(1\right) \\ \mathrm{Putting}x=0\mathrm{in}\mathrm{eq}\left(1\right) \\ \Rightarrow 1=A\left(0-2\right)\left(0-4\right)+B\times 0+C\times 0 \\ \Rightarrow \frac{1}{8}=A \\ \mathrm{Putting}\left(x-2\right)=0\mathrm{or}x=2\mathrm{in}\mathrm{eq}\left(1\right) \\ \Rightarrow 1=A\times 0+B\left(2\right)\left(2-4\right)+C\times 2\times 0 \\ \Rightarrow B=-\frac{1}{4} \\ \mathrm{Putting}\left(x-4\right)=0\mathrm{or}x=4\mathrm{in}\mathrm{eq}\left(1\right) \\ \Rightarrow 1=A\times 0+B\times 0+C\mathit{·}4\left(4-2\right) \\ \Rightarrow C=\frac{1}{8} \\ \therefore \frac{1}{x\left(x-2\right)\left(x-4\right)}=\frac{1}{8x}-\frac{1}{4\left(x-2\right)}+\frac{1}{8\left(x-4\right)} \\ \Rightarrow \int \frac{dx}{x\left(x-2\right)\left(x-4\right)}=\frac{1}{8}\int \frac{1}{x}dx-\frac{1}{4}\int \frac{1}{x-2}dx+\frac{1}{8}\int \frac{1}{x-4}dx \\ =\frac{1}{8}\ln \left|x\right|-\frac{1}{4}\ln \left|x-2\right|+\frac{1}{8}\ln \left|x-4\right|+C \\ =\frac{1}{8}\left(\ln \left|x\right|+\ln \left|x-4\right|-2\ln \left|x-2\right|\right)+C \\ =\frac{1}{8}\left[\ln \left|\frac{x\left(x-4\right)}{{\left(x-2\right)}^2}\right|\right]+C \end{gathered}$$