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