`int((4e^x-25)/(2e^x-5))dx=A x+B log/(2e^x)-5/(+c)` then

Options

A. A = 5 and B = 3
B. A = 5 and B = -3
C. A = -5 and B = 3
D. A = -5 and B = -3

Answer

B

Explanation

Let `l=int((4e^(x)-25)/(2e^(x)-5))dx`
`=int(4e^(x))/(2e^(x)-5)dx-int(25)/(2e^(x)-5)dx`
`=4int(e^(x))/(2e^(x)-5)dx-25int(e^(-x))/(2-5e^(-x))dx`
Put`" "2e^(x)-5=u and 2-5e^(-x)=v`
`rArr" "2e^(x)dx=du and 5e^(-x)dx=dv`
`rArr" "E^(x)dx=(du)/(2) and e^(-x)dx=(dv)/(5)`
`therefore" "l=4int(du)/(2u)-25int(Du)/(5v)`
`=2log u-5logv+c`
`=2log(2e^(x)-5)-5log(2-5e^(-x))+c`
`=2log(2e^(x)-5)-5log((2e^(x)-5)/(e^(x)))+c`
`=2log(2e^(x)-5)-5log((2e^(x)-5)/(e^(x)))+c`
`=2log(2e^(x)-5)-5log (2e^(x)-5)+5loge^(x)+c`
`=-3log(2e^(x)-5)+5x+c`
`rArr" "I=5x-3log(2e^(x)-5)+c`
But it is given `I=Ax+B log (2e^(x)-5)+c`
`therefore" A"=5 and B=-3`