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

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#### Solution

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

Putting e^{x} = t and e^{x}dx = 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 t^{2} on both sides of (1), we get

*B + C = 0*

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