Easiest way to take the integral of(involving substitution)

MelissaJL
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What is the easiest way to take the integral of:
\int\frac{(6+e^{x})^{2}dx}{e^{x}}

I have been having quite some difficulties with this one but here is my work so far:

Let u=e^{x}, du=e^{x}dx
=\int\frac{(u+6)^{2}du}{u^{2}}
Then let s=u+6 ∴ u=s-6, ds=du
=\int\frac{s^{2}ds}{(s-6)^{2}}
=\int\frac{s^{2}ds}{36-12s+s^{2}}

At this point I find myself lost and am not sure what to do. Is there an easier way to solve this integral? Also, if this is the only way to solve it how do I finish it from here? Thank you :)
 
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MelissaJL said:
What is the easiest way to take the integral of:
\int\frac{(6+e^{x})^{2}dx}{e^{x}}

I have been having quite some difficulties with this one but here is my work so far:

Let u=e^{x}, du=e^{x}dx
=\int\frac{(u+6)^{2}du}{u^{2}}
Then let s=u+6 ∴ u=s-6, ds=du
=\int\frac{s^{2}ds}{(s-6)^{2}}
=\int\frac{s^{2}ds}{36-12s+s^{2}}

At this point I find myself lost and am not sure what to do. Is there an easier way to solve this integral? Also, if this is the only way to solve it how do I finish it from here? Thank you :)


I think the easiest way to do it is just multiply out (6+e^x)^2.
 
Dick said:
I think the easiest way to do it is just multiply out (6+e^x)^2.

Haha, I don't know why I didn't just do that, apparently I want to just make things more complicated than they really are...
So then multiplying out gives me:
\frac{(6+e^{x})^{2}}{e^{x}} = \frac{e^{2x}+12e^{x}+36}{e^{x}}
\int (ex+12+36e-x)dx
= ex+12x-36e-x+C


Thank you.
 
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