# Integral of e to the Power of a Function

The integration of $$e$$ to the power $$x$$ of a function is a general formula of exponential functions and this formula needs a derivative of the given function. This formula is important in integral calculus.

The integration of e to the power x of a function is of the form

\[\int {{e^{f\left( x \right)}}f’\left( x \right)dx = } {e^{f\left( x \right)}} + c\]

Now consider

\[\frac{d}{{dx}}\left[ {{e^{f\left( x \right)}} + c} \right] = \frac{d}{{dx}}{e^{f\left( x \right)}} + \frac{d}{{dx}}c\]

Using the derivative formula $$\frac{d}{{dx}}{e^{f\left( x \right)}} = {e^{f\left( x \right)}}f’\left( x \right)$$, we have

\[\begin{gathered} \frac{d}{{dx}}\left[ {{e^{f\left( x \right)}} + c} \right] = {e^{f\left( x \right)}}f’\left( x \right) + 0 \\ \Rightarrow {e^{f\left( x \right)}}f’\left( x \right) = \frac{d}{{dx}}\left[ {{e^{f\left( x \right)}} + c} \right] \\ \Rightarrow {e^{f\left( x \right)}}f’\left( x \right)dx = d\left[ {{e^{f\left( x \right)}} + c} \right] \\ \end{gathered} \]

Integrating both sides of equation (i) with respect to $$x$$, we have

\[\int {{e^{f\left( x \right)}}f’\left( x \right)dx} = \int {d\left[ {{e^{f\left( x \right)}} + c} \right]} \]

Since integration and differentiation are reverse processes to each other, the integral sign $$\int {} $$ and $$\frac{d}{{dx}}$$ on the right side will cancel each other out, i.e.

\[\int {{e^{f\left( x \right)}}f’\left( x \right)dx = } {e^{f\left( x \right)}} + c\]

__Example__**:** Evaluate the integral $$\int {\frac{{{e^{{{\sin }^{ – 1}}x}}}}{{\sqrt {1 – {x^2}} }}dx} $$ with respect to $$x$$

We have integral \[I = \int {\frac{{{e^{{{\sin }^{ – 1}}x}}}}{{\sqrt {1 – {x^2}} }}dx} \]

Here $$f\left( x \right) = {\sin ^{ – 1}}x$$, and $$d\left( x \right) = \frac{1}{{\sqrt {1 – {x^2}} }}$$, so we can write it as

\[\int {\frac{{{e^{{{\sin }^{ – 1}}x}}}}{{\sqrt {1 – {x^2}} }}dx} = \int {{e^{{{\sin }^{ – 1}}x}}\frac{1}{{\sqrt {1 – {x^2}} }}dx} \]

Using the integration formula $$\int {{e^{f\left( x \right)}}f’\left( x \right)dx = } {e^{f\left( x \right)}} + c$$, we have

\[\int {\frac{{{e^{{{\sin }^{ – 1}}x}}}}{{\sqrt {1 – {x^2}} }}dx} = {e^{{{\sin }^{ – 1}}x}} + c\]

Dave Hysom

February 21@ 8:47 pmmathematic explanations without intuition is meaningless for the general public. This is why people do not like math. Mathematics would be much more popular — as would be scientists and academics — it they included intuition in their explanations. Also, any internet post that does not include a date should be consigned to the outer reaches of hell. If you are absolutely going to follow the deplorable practice of omitting intuition, you should clearly label your post, “this will only make sense if you already understand what I’m talking about.” Thanks for listening!