Evaluate : `int(6x+7)/(sqrt((x-5)(x-4)))\ dx`

Explanation

`" Let " I= int (6x+7)/(sqrt((x-5)(x-4)))dx`
` = int(6x+7)/(sqrt(x^(2)-9x+20))dx`
`Let 6x+7 =A .(d)/(dx)(x^(2)-9x+20)+B`
`rArr 6x+7 =A (2x-9)+B`
Comparing the coefficient of x,
`6=2A rArr A=3`
Comparing constant terms
`7=- 9A+B`
`rArr 7=-27 +B`
`rArr B=34`
`:. I= int (3(2x-9)+34)/(sqrt(x^(2)-9x+20))dx`
`=3 int (2x-9)/(sqrt(x^(2)-9x+20))dx" "underset(rArr (2x-9) dx=dt)underset(" Let "x^(2) -9x +20=t)(" For first intergral")`
`+34 int (1)/(sqrt(x^(2)-9x+20))dx`
`=3int (1)/(sqrt(t))dt+34 int (1)/(sqrt((x^(2)-9x+(81)/(4))+20-(82)/(4)))dx`
`=3 int t^(-1//2) dt+34 int(1)/(sqrt((x-(9)/(2))^(2)-((1)/(2))^(2)))dx`
`=(3.t^(1//2))/(1//2)+34 log |x-(9)/(2)+sqrt((x-(9)/(2))^(2)-((1)/(2))^(2))|+c`
`=6sqrt(t)+34log |x-(9)/(2)+sqrt(x^(2)-9x+20)|+c`
`=6sqrt(x^(2)-9x+20)`
`+34 log |x-(9)/(2)+sqrt(x^(2)-9x+20)|+c`