Question

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

Answer 1

`int 3^(2x)/(1+3^(2x))dx=1/(2ln3)ln abs(1+3^(2x))+C`

Explanation:

Substitute `t=3^(2x)=e^(2ln3x)`, `dt= 2ln3*e^(2ln3x)`

`int 3^(2x)/(1+3^(2x))dx = 1/(2ln3)int (dt)/(1+t)= 1/(2ln3)ln abs(1+t)+C=1/(2ln3)ln abs(1+3^(2x))+C`

Answer 2

Do a u substitution. Please see the explanation.

Explanation:

`int(3^(2x))/(1 + 3^(2x))dx =`

`int9^x/(1 + 9^x)dx`

Let `u = 1 + 9^x`, then `(du)/dx = (d(1))/dx + (d(9^x))/dx`

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

let `y = 9^x`, then `(du)/dx = 0 + dy/dx`

`ln(y) = ln(9^x)`

`ln(y) = xln(9)`

`1/ydy/dx = ln(9)`

`dy/dx = ln(9)y`

`dy/dx = ln(9)9^x`

Substituting this into the equation for `(du)/dx`

`(du)/dx = ln(9)9^x`

Writing this so that we can substitute into the integral:

`9^xdx = 1/ln(9)du`

Substituting this and u into the integral:

`1/ln(9)int(du)/u = 1/ln(9)ln(u) + C`

Reverse the substitution for u:

`int(3^(2x))/(1 + 3^(2x))dx = 1/ln(9)ln(1 + 9^x) + C`