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Question Number 51420 by Tawa1 last updated on 26/Dec/18

$$\int \frac{{\mathrm{e}}^{3\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}$$

Answered by ajfour last updated on 26/Dec/18

$$let1+e^x=t,\Rightarrow e^{x}dx=dt$$$$and{e}^{3x}dx=e^{2x}\left(e^{x}dx\right)={(t-1)}^{2}dt$$$$I=\int \frac{{(t-1)}^{2}dt}{t}=\int (t-2+\frac{1}{t})dt$$$$\Rightarrow I=\frac{{(1+e^x)}^2}{2}-2e^x+\ln (1+e^x)+c.$$

Commented by Tawa1 last updated on 26/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by Smail last updated on 26/Dec/18

$$lett=e^x\Rightarrow e^{x}dx=dt$$$$I=\int \frac{e^{3x}}{1+e^x}dx=\int \frac{t^2}{1+t}dt=\int \frac{t(t+1)-(t+1)+1}{1+t}dt$$$$=\int (t-1+\frac{1}{1+t})dt$$$$=\frac{t^2}{2}-t+ln\mid t+1\mid +C$$$$I=\frac{e^{2x}}{2}-e^x+ln(e^x+1)+C$$

Commented by Tawa1 last updated on 26/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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