Question

Evaluate : $\int \frac{1-2x^2}{x^3-3x}dx$

Answer 1

$\int \frac{1-{2x}^2}{x^3-3x}=\int \frac{1-{2x}^2}{x(x^2-3)}=\int \frac{1-{2x}^2}{x(x-\sqrt{3})(x+\sqrt{3})}$

$\Rightarrow \frac{1-{2x}^2}{(x^3-3x)}=\frac{A}{x}+\frac{B}{(x-\sqrt{3})}+\frac{C}{x+\sqrt{3}}$

$\Rightarrow 1-2x^2=A(x^2-3)+Bx(x+\sqrt{3})+Cx(x-\sqrt{3})$

$=A{x}^2-3A+B{x}^2+\sqrt{3}Bx+C{x}^2-\sqrt{3}Cx$

$\Rightarrow 1=-3A$

$0=\sqrt{3}B-C\sqrt{3}$

$-2=A+B+C$

$\sqrt{3}B=\sqrt{3}C$

$\Rightarrow B=C$

$\Rightarrow A=\frac{-1}{3}$

$-2=A+B+C$

$-2=\frac{-1}{3}+B+B$

$-2+\frac{1}{3}=2B$

$\Rightarrow B=\frac{-5}{6}$

$\Rightarrow c=\frac{-5}{6}$

$\Rightarrow \int \frac{1-{2x}^2}{x^3-3x}=\frac{-1}{3}\int \frac{1}{x}dx-\frac{5}{6}\int \frac{1}{x+\sqrt{3}}dx$

$=\frac{-1}{3}\log x-\frac{5}{6}\log (x-\sqrt{3})-\frac{5}{6}\log (x+\sqrt{3})$

$=\frac{-1}{3}\log x-\frac{5}{6}[\log (x-\sqrt{3})+\log (x+\sqrt{3})]$

$=-\frac{1}{3}\log x-\frac{5}{6}\log (x^2-3)+c$

Hence, the answer is $-\frac{1}{3}\log x-\frac{5}{6}\log (x^2-3)+c.$