`(4e^(x)-25)/(2e^(x)-5)`

Explanation

`"Let I"=int (4e^(x)-25)/(2e^(x)-5)dx`
Put Numerator=A(Denominator)+`B[(d)/(dx)("Denominator")]`
`therefore 4e^(x)-25=A(2e^(x)-5)+B[(d)/(dx)(2e^(x)-5)]`
`=A(2e^(x)-5)+B(2e^(x)-0)`
`therefore 4e^(x)-25=(2A+2B)e^(x)-5A`
Equation the coefficient of `e^(x)` and constant on both sides, we get
`2A+2B=4 ......(1)`
and 5A=25 `therefore A=5`
`therefore "from "(1),2(5)+2B=4`
`therefore 2b=-6 " "therefore B=-3`
`therefore 4e^(x)-25=5(2e^(x)-5)-3(2e^(x))`
`therefore I=int [(5(2e^(x)-5)-3(2e^(x)))/(2e^(x)-5)]dx`
`=int [5-(3(2e^(x)))/(2e^(x)-5)]dx`
`=5int Idx-3 int(2e^(x))/(2e^(x)-5)dx`
`=5x-3log |2e^(x)-5|+c" ".....[therefore int (f'(x))/(f(x))dx=log|f(x)|+c]`