Re: HSC 2017 MX2 Integration Marathon
Let\quad u=\frac { (k+1)x }{ x+k } \\ du=\frac { k(k+1)\quad dx }{ { (x+k) }^{ 2 } } \\ x=0,\quad u=0\\ x=1,\quad u=1\\ \frac { 1-x }{ x+k } =\frac { 1-u }{ k } \\ \frac { x }{ x+k } =\frac { u }{ k+1 } \\ \therefore \int _{ 0 }^{ 1 }{ \frac { { x }^{ a-1...