Question

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

Options

• A = 5 and B = 3

• A = 5 and B = –3

• A = –5 and B = 3

• A = –5 and B = –3

Answer 1

`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then A = 5 and B = –3.

Explanation:

Consider, I = `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`

Let 2e^x – 5 = u and 2 – 5e^–x = v

`\implies` 2e^xdx = du and 5e^–x dx = dv

`\implies` e^xdx = `(du)/2` and e^–x dx = `(dv)/5`

So, I = `4int (du)/(2u) - 25int (du)/(5v)`

= 2 log u – 5 log v + c

= 2 log(2e^x – 5) – 5 log (2 – 5e^–x) + c

= `2 log(2e^x - 5) - 5 log((2e^x - 5)/e^x) + c`

= –3 log (2e^x – 5) + 5x + c

Therefore, I = 5x – 3 log(2e^x – 5) + c

As, I = Ax + B log(2e^x – 5) + c

Hence, A = 5 and B = –3