# How do you integrate f(x)=e^(3x)/(1+e^x) using the quotient rule?

Oct 29, 2017

$I = \int \frac{{e}^{3 x} \cdot \mathrm{dx}}{1 + {e}^{x}} = \frac{1}{2} \cdot {e}^{2 x} - {e}^{x} + L n \left({e}^{x} + 1\right) + C$

#### Explanation:

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

=$\int \frac{{e}^{3 x} \cdot \left({e}^{x} + 1\right) \cdot \mathrm{dx}}{1 + {e}^{x}} ^ 2$

=$\int \left({e}^{3 x} + {e}^{2 x}\right) \cdot \frac{{e}^{x} \cdot \mathrm{dx}}{1 + {e}^{x}} ^ 2$

=$\left({e}^{3 x} + {e}^{2 x}\right) \cdot - \frac{1}{{e}^{x} + 1}$+$\int \frac{\left(3 {e}^{3 x} + 2 {e}^{2 x}\right) \cdot \mathrm{dx}}{1 + {e}^{x}} - 2 C$

=$- \frac{{e}^{3 x} + {e}^{2 x}}{{e}^{x} + 1}$+$3 \cdot \int \frac{{e}^{3 x} \cdot \mathrm{dx}}{1 + {e}^{x}}$+int (2e^(2x)*dx)/(1+e^x)-2C

=$- \frac{{e}^{2 x} \cdot \left({e}^{x} + 1\right)}{{e}^{x} + 1}$+$3 I$+$\int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{1 + {e}^{x}} - 2 C$

=$- {e}^{2 x} + 3 I$+$\int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{1 + {e}^{x}} - 2 C$

Hence $- 2 I$=$- {e}^{2 x}$+$\int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{1 + {e}^{x}}$-$2 C$

$J = \int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{1 + {e}^{x}}$

=$\int \frac{2 {e}^{2 x} \cdot \left({e}^{x} + 1\right) \cdot \mathrm{dx}}{1 + {e}^{x}} ^ 2$

=$\int \left(2 {e}^{2 x} + 2 {e}^{x}\right) \cdot \frac{{e}^{x} \cdot \mathrm{dx}}{1 + {e}^{x}} ^ 2$

=$\left(2 {e}^{2 x} + 2 {e}^{x}\right) \cdot - \frac{1}{{e}^{x} + 1}$+$\int \frac{\left(4 {e}^{2 x} + 2 {e}^{x}\right) \cdot \mathrm{dx}}{{e}^{x} + 1}$

=$- \frac{2 {e}^{2 x} + 2 {e}^{x}}{{e}^{x} + 1}$+$2 \cdot \int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{{e}^{x} + 1}$+$2 \cdot \int \frac{{e}^{x} \cdot \mathrm{dx}}{{e}^{x} + 1}$

=$\frac{- 2 {e}^{x} \cdot \left({e}^{x} + 1\right)}{{e}^{x} + 1} + 2 J + 2 L n \left({e}^{x} + 1\right)$

=$- 2 {e}^{x} + 2 J + 2 L n \left({e}^{x} + 1\right)$

So,

$- J = - 2 {e}^{x} + 2 L n \left({e}^{x} + 1\right)$ or,

$J = 2 {e}^{x} - 2 L n \left({e}^{x} + 1\right)$

Thus,

$- 2 I$=$- {e}^{2 x} + J - 2 C$

=$- {e}^{2 x} + 2 {e}^{x} - 2 L n \left({e}^{x} + 1\right) - 2 C$ or,

$I$=$\frac{1}{2} \cdot {e}^{2 x} - {e}^{x} + L n \left({e}^{x} + 1\right) + C$