Evaluate the indefinite integral
∫3^2x/1+3^2x dx

Question

Evaluate the indefinite integral
 ∫3^2x/1+3^2x dx

anonymous · Answer

@mgeorge20 let u = 1 + 3^2x

anonymous · Answer

@mgeorge20 since the base of (u) is a constant to find du we use (rule # 12)

$$\frac{ d }{ dx } [a^{u}] = (\ln(a))a ^{u}u'$$

anonymous · Answer

I should have mentioned first let's re-write it as 

$$\int\limits_{?}^{?}\frac{ 1 }{ 1 + 3^{2x} } * 3^{2x}dx$$

to solve du of 3^2x

a = 3

Let's use a different variable for 2x

v = 2x

du = ln(3)(3^2x)(2)

we want du should equal 3^2x

so that we could write an equation

$$\int\limits_{?}^{?}\frac{ 1 }{ u } du$$

anonymous · Answer

@mgeorge20  the ? is not supposed to be there

anonymous · Answer

du = ln(3)(3^2x)(2) dx

anonymous · Answer

If we divide both sides by 2 ln(3) we get 

du/[2 ln(3)] = 3^2x dx

anonymous · Answer

@mgeorge20

$$\int\limits_{}^{} \frac{ 1 }{ u } * \frac{ du }{ 2 \ln(3) }$$

2 ln(3) is a constant so we can put in front

$$\frac{ 1 }{ 2 \ln(3) } \int\limits_{}^{}\frac{ 1 }{ u }*du$$

anonymous · Answer

@mgeorge20 let me know when you're ready for the next step

solomonzelman · Answer

$\large\color{#006699}{ \displaystyle \int\frac{a^{x} }{b+a^{x}}{~\rm d}x }$


$\large\color{#039699}{ \displaystyle u=b+a^{x} }$

$\large\color{#039699}{ \displaystyle a^{x}=u-b }$

$\large\color{#039699}{ \displaystyle du/dx= \ln
(a)\cdot a^{x} }$

$\large\color{#039699}{ \displaystyle du= \ln
(a)\cdot a^{x}dx }$

$\large\color{#039699}{ \displaystyle du= \ln
(a)\cdot (u-b)dx }$

$\large\color{#039699}{ \displaystyle \frac{1}{\ln(a)}\frac{1}{ (u-b)}du= dx }$


$\large\color{#006699}{ \displaystyle\frac{1}{\ln(a)}\cdot \int\frac{u-b}{u} \cdot\frac{1}{ (u-b)}{~\rm d}u }$

$\large\color{#006699}{ \displaystyle\frac{1}{\ln(a)}\cdot \int\frac{1}{u} {~\rm d}u }$

$\large\color{#006699}{ \displaystyle\frac{1}{\ln(a)}\ln(u)\color{grey}{+C} }$

$\large\color{#006699}{ \displaystyle\frac{1}{\ln(a)}\ln(b+a^{x})\color{grey}{+C} }$

solomonzelman · Answer

$\large\color{#222299}{ \displaystyle \int\frac{a^{x} }{b+a^{x}}{~\rm d}x =\frac{1}{\ln(a)}\ln(b+a^{x})\color{grey}{+C} }$

anonymous · Answer

@LeibyStrauss go ahead

anonymous · Answer

$$\int\limits_{}^{}\frac{ 1 }{ u } = \ln \left| u \right|+ c$$

solomonzelman · Answer

(Yes.... assuming we don't get into imaginaries, the u is positie for all x anyway.

anonymous · Answer

$$\frac{ 1 }{ 2 \ln(3) } * \ln \left| 1 + 3^{2x} \right|+ C$$

anonymous · Answer

Previous post is the final answer