Another approach - a to u Integral

[Jason76]

\(\displaystyle \int 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2 dx\)

\(\displaystyle \dfrac{1}{2} \int 3^{u} du\)

\(\displaystyle \dfrac{3^{u}}{2} + C\)

 

[evinda]

\(\displaystyle \int 3^{2x}dx=\int 3^{u}\frac{du}{2} (1), u=2x=>dx=\frac{du}{2}.


(1)=>\frac{1}{2}\int 3^{u}du=\frac{1}{2ln3}3^{u}+c \)

 

[Deleted member 4993]

\(\displaystyle \int 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2 dx\)

\(\displaystyle \dfrac{1}{2} \int 3^{u} du\)

\(\displaystyle \dfrac{3^{u}}{2} + C\) ← This does not follow from above

.

 

[DrPhil]

please go back and read the beginning of the previous thread. That is the substitution we used to begin with.

 

[Jason76]

please go back and read the beginning of the previous thread. That is the substitution we used to begin with.

I know how to use that method. However, I also want to know the other method.

\(\displaystyle \int 3^{2x}dx=\int 3^{u}\frac{du}{2} (1), u=2x=>dx=\frac{du}{2}.


(1)=>\frac{1}{2}\int 3^{u}du=\frac{1}{2ln3}3^{u}+c \)

That makes sense. However, I would still probably use the method that Dr. Phil and others advise.