Question

Integrate the following w.r.t. x: ${\displaystyle \frac{2x^2-1}{x^4+9x^2+20}}$

Answer 1

Let I = ${\displaystyle \int \frac{2x^2-1}{x^4+9x^2+20}.dx}$

Consider, ${\displaystyle \frac{2x^2-1}{x^4+9x^2+20}}$

For finding partial fractions only, put x² = t.

∴ ${\displaystyle \frac{2x^2-1}{x^4+9x^2+20}=\frac{t}{(t+1)(t-2)(t+3)}}$

= ${\displaystyle \frac{\text{A}}{t+1}+\frac{\text{B}}{t-2}+\frac{\text{C}}{t+3}}$             ...(Say)

∴ t = A(t – 2)(t + 3) + B(t + 1)(t + 3) + C(t + 1)(t –2)

Put t + 1 = 0, i.e. t = – 1, we get

–1 = A(– 3)(2) + B(0)(2) + C(0)(– 3)

∴ – 1 = – 6A

∴ A = ${\displaystyle \frac{1}{6}}$

Put t – 2 = 0, i.e. t = 2, we get

2 = A(0)(5) + B(3)(5) + C(3)(0)

∴ 2 = 15B

∴ B = ${\displaystyle \frac{2}{15}}$

Put t + 3 = 0, i.e. t = – 3, we get

– 3 = A(–  5)(0) + B(–  2)(0) + C(– 2)(–  5)

–3 = 10C

∴ C = ${\displaystyle -\frac{3}{10}}$

∴ ${\displaystyle \frac{t}{(t+1)(t-2)(t+3)}=\frac{\left(\frac{1}{6}\right)}{t+1}+\frac{\left(\frac{2}{15}\right)}{x^2-2}+\frac{\left(\frac{-3}{10}\right)}{x^2+3}}$

∴ ${\displaystyle \frac{x^2}{(x^2+1)(x^2-2)(x^2+3)}=\frac{\left(\frac{1}{6}\right)}{x^2+1}+\frac{\left(\frac{2}{15}\right)}{x^2-2}+\frac{\left(\frac{-3}{10}\right)}{x^2+3}}$

∴ I = ${\displaystyle \int [\frac{\left(\frac{1}{6}\right)}{x^2+1}+\frac{\left(\frac{2}{15}\right)}{x^2-2}+\frac{\left(\frac{-3}{10}\right)}{x^2+3}].dx}$

= ${\displaystyle \frac{1}{6}\int \frac{1}{1+x^2}.dx+\frac{2}{15}\int \frac{1}{x^2-{\left(\sqrt{2}\right)}^2}.dx-\frac{3}{10}\int \frac{1}{x^2+{\left(\sqrt{3}\right)}^2}.dx}$

= ${\displaystyle \frac{1}{6}{\tan }^{-1}x+\frac{2}{15}\times \frac{1}{2\sqrt{2}}\log \left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|-\frac{3}{10}\times \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+c}$

= ${\displaystyle \frac{11}{\sqrt{5}}{\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)-\frac{9}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+c}$.