Integral of e to the Power of a Function

Integration of e power x of some function is a general formula of exponential function and this formula need derivative of the given function and this formula has an importance in integral calculus.

The integration of e power x of 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 , so the integral sign \int  {} and \frac{d}{{dx}} on the right side will cancel each other, 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 implies that d\left(  x \right) = \frac{1}{{\sqrt {1 - {x^2}} }}, so using formula, we have

\int  {\frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}dx}  = \int {{e^{{{\sin }^{ - 1}}x}}\frac{1}{{\sqrt  {1 - {x^2}} }}dx}

Using 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