Integrating $3^{2x}$ ($\int 3^{2x} dx $)

Question

I am tring to integrate the following: $3^{2x}$

My working has been shown as the following

$$\int 3^{2x} dx $$

I used the substitution $u = 2x$, so

$$\frac{du}{2} = dx $$

hence

$$\int 3^{2x} dx = \int 3^{u} \frac{du}{2} = \frac{1}{2}\int 3^{u} du $$

However, I can't go any further.

From this, I can't seem to prove the general result of $\int a^{x} dx $ either.

Can someone help?

Thanks!

Answer 1

Note that $3^{2x} = 9^x$, so you are basically back to square one.

Next, note that $9^x = e^{x \ln 9}$. To see this : $$ e^{x \ln 9} = (e^{\ln 9})^x = 9^x $$

Hence, $$ \int 3^{2x} dx = \int 9^x dx = \int e^{x \ln 9} dx $$ Now, let $y = x \ln 9$, so that $dy = (\ln 9) dx$. $$ \int e^{x \ln 9} dx = \int e^y \frac {dy}{\ln 9} = \frac{e^y}{\ln 9} + C = \frac{9^x}{\ln 9} + C $$

It's worth noticing that the derivative is the same as the original function up to a constant ($\frac 1{\ln 9}$) and the constant of integration.

Answer 2

Hint: $\displaystyle\int{a^{mx}}\ dx=\dfrac{a^{mx}}{m\cdot\ln a}+c$

Answer 3

Formula -

$$\int a^x = \frac{a^x}{\ln a} + c$$

Now we have,

$$\frac{1}{2}\int 3^{u} du $$

$$= \frac{1}{2} \cdot \frac{3^u}{\ln 3} + c$$

$$= \frac{1}{2} \cdot \frac{3^{2x}}{\ln 3} + c$$

Answer 4

Note that $$ D3^{u} = D(\exp (u\log 3)) = \exp(u\log 3)\cdot \log 3 = 3^{u}\log 3. $$ So $$ \int 3^{u} = \int \frac{D3^{u}}{\log 3} = \frac{3^{u}}{\log 3} + \text{constant}. $$