Can I Integrate E-sign in Jitterbit

Can I Integrate E-sign in Jitterbit. SignNow integrations bring more benefits for your business workflow. Create and manage documents, add and gather signatures without leaving your personal account. Secure and simple!

Easy Way to Integrate E-sign in Jitterbit

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How To Integrate E-sign in Jitterbit

in this video we're going to focus on integrating exponential functions so let's go ahead and begin what is the antiderivative of e 3x DX we think the answer is for this function it's simply going to equal the same thing it's e to the X plus C now what about this one what is the antiderivative of e to the 2x DX this is going to be e to the 2x divided by the derivative of 2x which is 2 plus C that technique works only if you divided by constant if you divided by a variable it's not going to work so what do you think the answer for this one is e to the negative 5x DX it turns out that it's going to be the same thing e to the negative 5x divided by the derivative of negative 5 X which is negative 5 so let's see now let's prove the last one using u substitution so let's make u equal to a negative 5 X D U is the derivative of negative 5 X which is negative 5 times DX and solving for DX it's D U divided by negative 5 X so now let's replace negative 5 X with the u variable and let's replace a DX with D u over negative 5 now let's take the constant negative 5 and let's move it to the front so it's negative 5 antiderivative of EU or e to the U vu now we know that the integration of e to the X is simply e to the X so the antiderivative of e to you is simply e to u plus C and now we can replace you with negative 5x to the final answer oh by the way this should be a negative 1 over 5 because the 5 is on the bottom so just keep that a month I almost forgot about that but the final answer is negative run over 5 e to the negative 5x plus C which is the same thing as what we have which was e to negative 5x divided by negative 5 plus C is another one X Q times e raised to be X to the fourth DX go ahead and integrate the function so we can integrate by a substitution let's replace U with X to the fourth so there for a BU it's going to be the derivative of X to the fourth which is 4x cubed times DX and that would solve for DX that's d u over 4x cubed so let's replace X to the fourth with the you variable and now let's replace DX with D u over 4x cubed and so we could cancel X cubed now let's move the fourth to the front so it's 1/4 antiderivative e to udu the integration of e to the U is just e to the U times 1/4 plus C and now all I need to do is replace U with X...


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