If ∫ sin - 1 x 1 + x d x = A x tan - 1 x + B x + C , where C is a constant of integration, then the ordered pair A x , B x can be :

Step-by-step solution

I = ∫ sin - 1 x 1 + x d x = ∫ tan - 1 x ⏟ I 1 ⏟ I I d x = x tan - 1 x - ∫ 1 1 + x . 1 2 x . x d x + C = x tan - 1 x - 1 2 ∫ t . 2 t . d t 1 + t 2 + C ( putting x = t 2 ⇒ d x = 2 t d t ) = x tan - 1 x - ∫ t 2 1 + t 2 d t + C = x tan - 1 x - t + tan - 1 t + C = x tan - 1 x - x + tan - 1 x + C = x + 1 tan - 1 x - x + C Comparing with A x tan - 1 x + B x + C , we get A x ,   B x = x + 1 ,   - x