How Do I Integrate eSign in Jitterbit

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Integration of e^(x^2)

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The integral in this question can be expressed in terms of what is called a special function.The series expansion of [math]\displaystyle e^{-x^2}[/math] around [math]x = 0[/math] is expressed as (verified with Mathematica):[math]\displaystyle 1-x^2 +\frac{x^4}{2}-\frac{x^6}{6} +\frac{x^8}{24}-\frac{x^{10}}{120}+ …[/math]The series expansion above is symbolically written as:[math]\displaystyle e^{-x^2}=\sum _{n=0}^{\infty } \frac{\left(-x^2\right)^n}{n!}=\sum _{n=0}^{\infty } \frac{(-1)^n \left(x^2\right)^n}{n!}[/math]Using the last result above and integrating, we get:[math]\displaystyle \int e^{-x^2} \, dx = \int \left(\sum _{n=0}^{\infty } \frac{(-1)^n \left(x^2\right)^n}{n!}\right) \, dx=\sum _{n=0}^{\infty } \frac{(-1)^n x^{2 n+1}}{(2 n+1) n!}+C[/math]If we examine the infinite sum obtained above, and look at the series expansion of the error function which is given by:[math]\displaystyle \operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{n!(2n+1)}} \\ \displaystyle \qquad ={\frac {2}{\sqrt {\pi }}}\left(x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{10}}-{\frac {x^{7}}{42}}+{\frac {x^{9}}{216}}-\cdots \right),[/math]we see that the integral in this question can be expressed as:[math]\displaystyle \boxed{\int e^{-x^2} \, dx =\sum _{n=0}^{\infty } \frac{(-1)^n x^{2 n+1}}{(2 n+1) n!}+C = \frac{1}{2} \sqrt{\pi }* \text{erf}(x)+C}[/math]Below is a plot of the integration solution:Note that the solution to the given indefinite integral could be found by using the definition of the error function:[math]\displaystyle \text{erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-x^{2}}\,dx={\frac {2}{\sqrt {\pi }}}\int e^{-x^{2}}\,dx,[/math]so that we have:[math]\displaystyle \int e^{-x^{2}}\,dx = \frac {\sqrt{\pi}}{2} \text{erf}(x)+ constant[/math]Considering the representation of the error function via other general functions, the given integral can be also expressed in terms of the generalized hypergeometric function [math]\, _pF_q (a;b;x)[/math] in the following form:[math]\displaystyle \int e^{-x^2} \, dx= x* \, _1F_1\left(\frac{1}{2};\frac{3}{2};-x^2\right) + constant [/math]

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[math]I=\displaystyle \int \dfrac 1 {e^{2x}+e^x}dx[/math][math]=\displaystyle \int \dfrac 1 {e^x(e^x+1)}dx[/math][math]=\displaystyle \int \dfrac {(e^x+1)-e^x} {e^x(e^x+1)}dx[/math][math]=\displaystyle \int \dfrac {1} {e^x}dx-\int \dfrac {1} {e^x+1}dx[/math][math]=\displaystyle \int e^{-x}dx-\int \dfrac {e^{-x}} {1+e^{-x}}dx[/math][math]=\boxed{-e^{-x}+\ln (1+e^{-x})+C}[/math]

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Letif y = integral e^x thendy/dx = e^xSupposey = f(x) = e^xSo derivative of e^x is e^x itself. or in other words integral of e^x is e^x itself