Integrate the following w. r. t. x : (x^2)/(x^2 + 1)(x^2​​​​​​​ - 2)(x^2​​​​​​​ + 3)

 (x²)/(x² + 1)(x² - 2)(x² + 3)

Let I = \(\int \frac{x^2}{(x^2 + 1)(x^2 - 2)(x^2 + 3)}dx\)

Consider, \(\frac{x^2}{(x^2 + 1)(x^2 - 2)(x^2 + 3)}\)

For finding partial fractions only, put x² = t.

\(\therefore\) \(\frac{x^2}{(x^2 + 1)(x^2 - 2)(x^2 + 3)}\) = \(\frac{t}{(t + 1)(t - 2)(t + 3)}\)

\(=\frac{A}{t + 1}+\frac{A}{t - 2}+\frac{A}{t + 3}\) ....(say)

\(\therefore\) 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)

\(\therefore\) -1 = -6A

\(\therefore\) A = 1/6

Put t - 2 = 0, i.e. t = 2, we get

2 = A(0)(5) + B(3)(5) + C(3)(0)

\(\therefore\) 2 = 15B

\(\therefore\) B = 2/15

Put t + 3 = 0, i.e. t = -3, we get

-3 = A(-5)(0) + B(-2)(0) + C(-2)(-5)

-3 = 10C

\(\therefore\) C \(=-\frac{3}{10}\)

\(\therefore\) \(\frac{t}{(t + 1)(t - 2)(t + 3)}\) \(=\frac{1/6}{t + 1}+\frac{2/15}{t - 2}+\frac{-3/10}{t + 3}\)

\(\therefore\) \(\frac{x^2}{(x^2 + 1)(x^2 - 2)(x^2 + 3)}\) \(=\frac{1/6}{x^2 + 1}+\frac{2/15}{x^2- 2}+\frac{-3/10}{x^2 + 3}\)

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

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

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

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