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 6x, 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 = 3x/2x
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 = 3x/2x, so
\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