# Thread: Simple but messed up

1. ## Simple but messed up

In my book the integral of a^x is directly given as $\displaystyle \int a^xdx=\frac{a^x}{log a}$
Can anyone tell me how can we derive it?

2. Use the identity $\displaystyle a^x=e^{\text{log}(a)x}$.

3. Originally Posted by roshanhero
In my book the integral of a^x is directly given as $\displaystyle \int a^xdx=\frac{a^x}{log a}$
Can anyone tell me how can we derive it?
let $\displaystyle y = a^x$, then $\displaystyle \ln y = x \ln a$ and $\displaystyle \frac{1}{y} \frac{dy}{dx} = \ln a$, or $\displaystyle dx = \frac{1}{y} \frac{dy}{\ln a}$

and the integral became

$\displaystyle \int a^xdx=\int y \frac{1}{y} \frac{dy}{\ln a} = \frac{a^x}{\ln a}$

4. Hi , First of All take your book away. it is not a mathematics , maybe it is

Literature book .

Solve The Integral By This Way :

$\displaystyle \int a^x dx = \int \left({e^{\ln a}}\right)^x dx = \int {{e^{(\ln a)x}}} dx$

$\displaystyle u=x\ln a \rightarrow du = dx\ln a$

$\displaystyle \int e^u \frac {du}{\ln a}=\ln a\int e^u = \frac {1}{\ln a} {a^x}$

Note :

$\displaystyle e^u = e^{x \ln a} =\left( e^{\ln a} \right)^{x} = a^x$

5. Originally Posted by roshanhero
In my book the integral of a^x is directly given as $\displaystyle \int a^xdx=\frac{a^x}{log a}$
Can anyone tell me how can we derive it?
You're expected to learn something from the help you get in other threads. In this instance I refer specifically to http://www.mathhelpforum.com/math-he...questions.html