Question

$\int \frac{x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}dx$

Answer 1

$$\begin{gathered} \int \frac{x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}dx \\ \mathrm{Let}\frac{x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ \Rightarrow \frac{x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{A\left(x-2\right)\left(x-3\right)+B\left(x-1\right)\left(x-3\right)+C\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)} \\ \Rightarrow x^2=A\left(x-2\right)\left(x-3\right)+B\left(x-1\right)\left(x-3\right)+C\left(x-1\right)\left(x-2\right)....\left(1\right) \\ \mathrm{Putting}x-1=0\mathrm{or}x=1\mathrm{in}\mathrm{eq}\left(1\right) \\ \Rightarrow 1=A\left(1-2\right)\left(1-3\right) \\ \Rightarrow 1=A\left(-1\right)\left(-2\right) \\ \Rightarrow A=\frac{1}{2} \\ \mathrm{Putting}x-2=0\mathrm{or}x=2\mathrm{in}\mathrm{eq}\left(1\right) \\ \Rightarrow 4=B\left(2-1\right)\left(2-3\right) \\ \Rightarrow B=-4 \\ \mathrm{Putting}x-3=0\mathrm{or}x=3\mathrm{in}\mathrm{eq}\left(1\right) \\ \Rightarrow 9=C\left(3-1\right)\left(3-2\right) \\ \Rightarrow C=\frac{9}{2} \\ \therefore \frac{x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{1}{2\left(x-1\right)}-\frac{4}{x-2}+\frac{9}{2\left(x-3\right)} \\ \int \frac{x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}dx=\frac{1}{2}\int \frac{1}{x-1}dx-4\int \frac{1}{x-2}dx+\frac{9}{2}\int \frac{1}{x-3}dx \\ =\frac{1}{2}\ln \left|x-1\right|-4\ln \left|x-2\right|+\frac{9}{2}\ln \left|x-3\right|+C \end{gathered}$$