If I=∫e3x1+exdx, then I equal

Put 1 + ex = t, so thatI=∫(t−1)2tdt=∫t−2+1tdt=12t2−2t+log⁡|t|+C =(1/2)1+exex−3+log⁡1+ex+C

If I=∫e3x1+exdx, then I equal

Put 1 + ex = t, so thatI=∫(t−1)2tdt=∫t−2+1tdt=12t2−2t+log⁡|t|+C =(1/2)1+exex−3+log⁡1+ex+C

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If $I = \int \frac{e^{3 x}}{1 + e^{x}} dx$ , then I equal

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$(1 / 2) (1 + e^{x}) (e^{x} - 3) + \log (1 + e^{x}) + C$

$(1 / 2) {(1 + e^{x})}^{2} - (1 / 3) (1 + e^{x}) + \log (1 + e^{x}) + C$

$(1 / 2) (1 + e^{x}) (e^{x} + 3) + \log (1 + e^{x}) + C$

$(1 / 2) {(1 + e^{x})}^{2} - 2 \log (1 + e^{x}) + C$

answer is A.

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Detailed Solution

Put 1 + e^x = t, so that

$\begin{array}{l} I = \int \frac{(t - 1)^{2}}{t} dt = \int (t - 2 + \frac{1}{t}) dt \\ = \frac{1}{2} t^{2} - 2 t + \log | t | + C = (1 / 2) (1 + e^{x}) (e^{x} - 3) + \log (1 + e^{x}) + C \end{array}$

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If $I = \int \frac{e^{3 x}}{1 + e^{x}} dx$ , then I equal