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Question
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 ex = t and exdx = 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
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