Question

`int (1 + 2e^-x)/(1 - 2e^-x)`dx = ?

Options

• x + log(1 - 2e^-x) + c

• x + 2 log(1 - 2e^-x) + c

• log(1 - 2e^-x) + c

• x - log(1 - 2e^-x) + c

Answer 1

x + 2 log(1 - 2e^-x) + c

Explanation:

We have,

I = `int (1 + 2e^-x)/(1 - 2e^-x)`dx

I = `int ((1 - 2e^-x)/(1 - 2e^-x)) "dx" + int(4"e"^-x)/(1 - 2"e"^-x)`dx

I = `int "dx" + 2 int "dt"/"t"`   ....[∵ put 1 - 2e^-x = t ⇒ 2e^-x dx = dt]

⇒ I = x + 2 log t + c

⇒ I = x + 2 log |1 - 2e^-x| + c