Integration by substitution

Integration by substitution

Postby Guest » Sun Jan 20, 2019 8:07 pm

I have a question where we need to find an integral using where "u = 1+[tex]e^{x}[/tex]" for the equation "[tex]\int \frac{e^{3x}}{1+e^{x}}[/tex]".

However when I substitute it I end up with "[tex]\int \frac{(u-1)^{3}}{u}[/tex]" instead of "[tex]\int \frac{(u-1)^{2}}{u}[/tex]" which is what I should be getting. please help
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Re: Integration by substitution

Postby HallsofIvy » Wed Mar 06, 2019 11:45 am

Your problem is that you have copied the original problem wrong! It is NOT [tex]\int \frac{e^{3x}}{1+ e^x}[/tex]. That doesn't make sense. It must be [tex]\int \frac{e^{3x}}{1+ e^x}dx[/tex]. You dropped the "dx" and that is crucially important!

With [tex]u= 1+ e^x[/tex], [tex]e^x= u- 1[/tex] so that [tex]e^{3x}= (e^x)^3= (u- 1)^3[/tex]. The denominator is, of course, u so that the fraction is [tex]\frac{(u- 1)^3}{u}[/tex] but since [tex]u= 1+ e^x[/tex], [tex]du= e^x dx[/tex], [tex]dx= \frac{1}{e^x} du= \frac{1}{u- 1}du[/tex].

So the integral becomes [tex]\int \frac{(u- 1)^3}{u}\left(\frac{1}{u- 1}du\right)= \int \frac{(u- 1)^2}{u} du[/tex]. That, of course, is easy to integrate.

HallsofIvy
 
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