evaluate the integral:

Question

int: [(x^3 + 3x^2 + x + 9)/(x^2 + 1)(x^2 + 3) ] dx

 

Answer 1

Given that (x³ + 3x² + x + 9) / [(x² + 1)(x² + 3)]

            = [(x³ + x) + (3x² + 9)] / [(x² + 1)(x² + 3)]

            = [x(x² + 1) + 3(x² + 3)] / [(x² + 1)(x² + 3)]

            = x(x² + 1) / [(x² + 1)(x² + 3)] + 3(x² + 3)] / [(x² + 1)(x² + 3)]

            = x / (x² + 3) + 3 / (x² + 1)

By applying integration we get,

ʃ(x / (x² + 3) + 3 / (x² + 1))dx = ʃxdx / (x² + 3) + 3ʃ1dx / (x² + 1)

                                          = 1/2 ʃ2xdx / (x² + 3) + 3ʃdx/ (x² + 1) [Divide and multiply by 2]

                                          = 1/2 ʃ 2xdx / (x² + 3) + 3 ʃ dx / (x² + 1)

[ Let x^2 + 3 = u  => du/dx = 2x  => du = 2xdx  ]

                                          = 1/2 ʃ du/u + 3tan^(-1)x + c             [Since ʃ dx / (x² + 1) = tan^(-1)x + c]

                                          = 1/2 logu + 3tan^(-1)x + c               [Since ʃ dx (1/x) = logx + c]

                                          = 1/2 log(x^2 + 3) + 3tan^(-1)x + c  where c is integration constant

Therefore the answer is 1/2 log(x^2 + 3) + 3tan^(-1)x + c.