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\(\int\frac{x+1}{x(1+xe^x)^2}dx\) |
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Answer» Let \(1 + xe^x = t\) Then \((xe^x + e^x)dx = dt\) ⇒ \(e^x (x + 1)dx = dt\) ⇒ \((x + 1) dx = \frac{dt}{e^x}\) \(\therefore \frac{(x + 1)dx}{x} = \frac{dt}{xe^x} = \frac{dt}{t -1}\) \(\therefore \int \frac{(x + 1) dx}{x(1 + xe^x)^2} = \int \frac{dt}{(t - 1)t^2}\) \(= \int \left(\frac{-1}{t} - \frac1{t^2} + \frac1{t -1}\right)dt\) \( = -log\, t + \frac1 t + log(t - 1) + C\) \(= log\left|\frac{t-1}{t}\right| + \frac1t + C\) \(= log\left|\frac{xe^x}{xe^x + 1}\right| + \frac1{xe^x + 1} + C\) |
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