Question

$\int \frac{x^2}{x^4-x^2-12}dx$

Answer 1

Let I =$\int \frac{x^2}{x^4-4x^2+3x^2-12}dx=\int \frac{x^{2}dx}{x^2(x^2-4)+3(x^2-4)}=\int \frac{x^{2}dx}{(x^2-4)(x^2+3)}Now,\frac{x^2}{(x^2-4)(x^2+3)}[let{x}^2=t]\Rightarrow \frac{t}{(t-4)(t+3)}=\frac{A}{t-4}+\frac{B}{t+3}\Rightarrow t=A(t+3)+B(t-4)Oncomparingthecoefficientoftonbothsides,wegetA+B=1\Rightarrow 3A-4B=0\Rightarrow 3(1-B)-4B=0\Rightarrow 3-3B-4B=07B=3\Rightarrow B=\frac{3}{7}$

If $B=\frac{3}{7},then,A+\frac{3}{7}=1\Rightarrow A=1-\frac{3}{7}=\frac{4}{7}\therefore I=\frac{4}{7}\int \frac{1}{x^2-(2{)}^2}dx+\frac{3}{7}\int \frac{1}{x^2+(\sqrt{3}{)}^2}dx=\frac{4}{7}.\frac{1}{2.2}log|\frac{x-2}{x+2}|+\frac{3}{7}.\frac{1}{\sqrt{3}}ta{n}^{-1}\frac{x}{\sqrt{3}}+C=\frac{1}{7}log|\frac{x-2}{x+2}|+\frac{\sqrt{3}}{7}ta{n}^{-1}\frac{x}{\sqrt{3}}+C$