How do you find the indefinite integral of int (3^(2x))/(1+3^(2x))?

Jan 7, 2017

$\int {3}^{2 x} / \left(1 + {3}^{2 x}\right) \mathrm{dx} = \frac{1}{2 \ln 3} \ln \left\mid 1 + {3}^{2 x} \right\mid + C$

Explanation:

Substitute $t = {3}^{2 x} = {e}^{2 \ln 3 x}$, $\mathrm{dt} = 2 \ln 3 \cdot {e}^{2 \ln 3 x}$

$\int {3}^{2 x} / \left(1 + {3}^{2 x}\right) \mathrm{dx} = \frac{1}{2 \ln 3} \int \frac{\mathrm{dt}}{1 + t} = \frac{1}{2 \ln 3} \ln \left\mid 1 + t \right\mid + C = \frac{1}{2 \ln 3} \ln \left\mid 1 + {3}^{2 x} \right\mid + C$

Jan 7, 2017

Do a u substitution. Please see the explanation.

Explanation:

$\int \frac{{3}^{2 x}}{1 + {3}^{2 x}} \mathrm{dx} =$

$\int {9}^{x} / \left(1 + {9}^{x}\right) \mathrm{dx}$

Let $u = 1 + {9}^{x}$, then $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d \left(1\right)}{\mathrm{dx}} + \frac{d \left({9}^{x}\right)}{\mathrm{dx}}$

The first term of the derivative is 0 but the second term requires logarithmic differentiation:

let $y = {9}^{x}$, then $\frac{\mathrm{du}}{\mathrm{dx}} = 0 + \frac{\mathrm{dy}}{\mathrm{dx}}$

$\ln \left(y\right) = \ln \left({9}^{x}\right)$

$\ln \left(y\right) = x \ln \left(9\right)$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(9\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(9\right) y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(9\right) {9}^{x}$

Substituting this into the equation for $\frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = \ln \left(9\right) {9}^{x}$

Writing this so that we can substitute into the integral:

${9}^{x} \mathrm{dx} = \frac{1}{\ln} \left(9\right) \mathrm{du}$

Substituting this and u into the integral:

$\frac{1}{\ln} \left(9\right) \int \frac{\mathrm{du}}{u} = \frac{1}{\ln} \left(9\right) \ln \left(u\right) + C$

Reverse the substitution for u:

$\int \frac{{3}^{2 x}}{1 + {3}^{2 x}} \mathrm{dx} = \frac{1}{\ln} \left(9\right) \ln \left(1 + {9}^{x}\right) + C$