Question

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

Answer 1

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

Putting e^x = t and e^xdx = dt, we get

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

Using partial fraction, we have

`1/((t-1)^2 (t + 1)) = A/(t-1)^2 +  B/(t -1) + C/(t +2)`

⇒ 1 = A(t+2) + B(t−1)(t+2) + C(t−1)² .....(1)

Putting t = 1 in (1), we get

`A = 1/3`

Putting t = −2 in (1), we get

C = `1/9`

Comparing the coefficients of t² on both sides of (1), we get

B + C = 0