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