# Find the Integral of an Exponential Function

# Indefinite integral Math problem

find the Indefinite Integral of an exponential function, 6 raise to x divided by (4 raised to x plus 9 raise to x)

\int{\dfrac{6^x}{4^x+9^x}} \ dx

## Solution: Integral of an exponential function

\mathrm{Let} \ \ \ \ I= \int{\dfrac{6^x}{4^x+9^x}} \ dx

Divide all terms with 6^{x}, then

I=\int{\dfrac{\dfrac{6^x}{6^x}}{\dfrac{4^x}{6^x}+\dfrac{9^x}{6^x}}} \ dx

\Rightarrow \ \ I=\int{\dfrac{1}{\dfrac{2^x}{3^x}+\dfrac{3^x}{2^x}}} \ dx

Substitute *u = *3^{x}/2^{x}

u=\dfrac{3^x}{2^x}

\Rightarrow \ \ \dfrac{du}{dx}=\dfrac{du}{dx}(\dfrac{3^x}{2^x})

\Rightarrow \ \ \dfrac{du}{dx}=\dfrac{2^x \times \dfrac{d(3^x)}{dx} -3^x \times \dfrac{d(2^x)}{dx} }{(2^x)^2}

\Rightarrow \ \ \dfrac{du}{dx}=\dfrac{ {2^x}. \ {3^x. \log3} -{3^x}. \ {2^x. \log2} }{2^{2x}}

\Rightarrow \ \ \dfrac{du}{dx}=\dfrac{{3^x. \log3} -{3^x}. \ {\log2} }{2^{x}}

\Rightarrow \ \ \dfrac{du}{dx}=\dfrac{3^x}{2^x} \times(\log3 - \log2)

We Know, *u = *3* ^{x}/*2

*, so*

^{x}\Rightarrow \ \ \dfrac{du}{dx}=u \times(\log3 - \log2)

dx=\dfrac{du}{u (\log3 - \log2)}

Now

I=\int{\dfrac{1}{\dfrac{2^x}{3^x}+\dfrac{3^x}{2^x}}} \ dx=\int{\dfrac{1}{\dfrac{1}{u}+u}} \times \dfrac{du}{u (\log3 - \log2)}

\Rightarrow \ \ I=\int{\dfrac{u}{1+u^2}} \times \dfrac{du}{u (\log3 - \log2)}

\Rightarrow \ \ I=\int{\dfrac{1}{1+u^2}} \times \dfrac{du}{\log3 - \log2}

\Rightarrow \ \ I=\dfrac{1}{\log3 - \log2}\int{\dfrac{du}{1+u^2}}

\Rightarrow \ \ I=\dfrac{1}{\log3 - \log2} \times \tan^{-1}u+C

Undo substitution

\Rightarrow \ \ I=\dfrac{1}{\log3 - \log2} \times \tan^{-1}(\dfrac{3^x}{2^x})+C

\Rightarrow \ \ I=\dfrac{\tan^{-1}(\dfrac{3^x}{2^x})}{\log3 - \log2}+C