If ∫ d x x 2 + x + 1 2 = a tan - 1 2 x + 1 3 + b 2 x + 1 x 2 + x + 1 + C , x > 0 where C is the constant of integration, then the value of 9 ( 3 a + b ) is equal to _________.

Step-by-step solution

I = ∫ d x x + 1 2 2 + 3 4 2 Put x + 1 2 = t = ∫ d t t 2 + 3 4 2     P u t     t = 3 2 tan   θ = 3 2 ∫ sec 2 θ   d θ 9 16 sec 4 θ         = 4 3 9 ∫ 1 + cos   2 θ d θ = 4 3 9 θ + sin   2 θ 2 + C = 4 3 9 tan - 1 2 x + 1 3 + 3 2 x + 1 3 + 2 x + 1 2 + C = 4 3 9 tan - 1 2 x + 1 3 + 1 3 2 x + 1 x 2 + x + 1 + C Hence, 9 3 a + b = 15