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