# Integration by substitution

### Integration by substitution

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

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

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

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

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

HallsofIvy

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