Evaluate the integral
\[ \int a^x \; dx \]
where \( a \) is a constant such that \( a \gt 0 \) and \( a \ne 1 \)
We first change the base of the exponential \( a^x \)
Let \( y = a^x \) and take the \( \ln \) of both sides
\[ \qquad \ln y = \ln (a^x) \]
The property of \( \; \ln \; \) given by \( \; \ln a^x = x \ln a \; \) in used to write the above as
\[ \qquad \ln y = x \ln a \]
Use the relationship between the logarithm and the exponential to write the above as
\[ y = e^{x \ln a} \]
and since \( y = a^x \), we obtain
\[ a^x = e^{x \ln a} \]
Using the above, the given integral may be written as
\[ \qquad \displaystyle \int a^x \; dx = \int e^{x \ln a} dx \qquad (1) \]
We now evaluate the integral on the right side
Let \( u(x) = x \ln a \) which gives \( \dfrac{du}{dx} = \ln a \)
By definition, 1 , 2 , 3 the differential \( du \) is given by \( du = \dfrac{du}{dx} dx = \ln a \; dx \), hence \( du = \ln a \; dx \)
Use the substitution \( u(x) = x \; \ln a \) in the right integral in \( (1) \) and write
\[ \displaystyle \int e^{x \ln a} dx = \int e^{u} \; dx \]
Divide and multiply the integrand on the right by \( \ln a \)
\[ \qquad \displaystyle \int e^{x \ln a} dx = \int \dfrac{1}{\ln a}e^{u} \ln a \; dx \]
Use the fact that \( du = \ln a \; dx \) in the above and rewrite the integral as
\[ \displaystyle \int e^{x \ln a} dx = \int \dfrac{1}{\ln a}e^{u} \; du \]
Evalute the above integral using the integral formula \( \displaystyle \int e^x dx = e^x + c \) to obtain
\[ \displaystyle \int e^{x \ln a} dx = \dfrac{1}{\ln a} e^u + c \]
where \( c \) is the constant on integration.
Substitute back \( u \) and write
\[ \displaystyle \int e^{x \ln a} dx = \dfrac{1}{\ln a}e^{x \ln a} + c \]
and finally using \( \displaystyle \int a^x \; dx = \int e^{x \ln a} dx \) in \( (1) \) above , we write
\[ \int a^x \; dx = \dfrac{1}{\ln a}e^{x \ln a} + c \]
or using the fact that \( a^x = e^{x \ln a} \),
\[ \int a^x \; dx = \dfrac{1}{\ln a} a^x + c \]