Ask your own question, for FREE!
Mathematics 68 Online
OpenStudy (anonymous):

They just keep getting more complex, help! Find the indefinite integral: Integrand of ((xe^(-x^2))+(e^x)/((e^x)+8))

OpenStudy (anonymous):

i don't even know where to begin!

OpenStudy (kinggeorge):

I would do one part at a time. Can you find \[\int\limits xe^{-x^2} dx\]yourself?

OpenStudy (anonymous):

um... u=(-x^2), du=-2x?

OpenStudy (kinggeorge):

looks good so far.

OpenStudy (anonymous):

so that part would be 1/2(e^(-x^2)+c?

OpenStudy (kinggeorge):

Put a negative in front, but otherwise, looks great.

OpenStudy (anonymous):

oh yeah..

OpenStudy (anonymous):

how do i determine u, for the second part - would it be (e^x)+8?

OpenStudy (kinggeorge):

That looks like the u that I would would use.

OpenStudy (anonymous):

now for du.. would it be just e^x?

OpenStudy (anonymous):

or actually be 1/(e^x)?

OpenStudy (kinggeorge):

Just \(e^x\)

OpenStudy (anonymous):

ok from there, would my substitution for the second part be: Integrand of du(u^-1)?

OpenStudy (anonymous):

and then just add them together as a final answer?

OpenStudy (kinggeorge):

Integrate\[\int\limits {1 \over u}\;\;du\]substitute the u back in, add them together, and put a little +C at the end.

OpenStudy (anonymous):

so the second part would = e^x(e^x+8)

OpenStudy (anonymous):

and add it to the first part.. hmmm

OpenStudy (kinggeorge):

I'm getting a different solution for the second part. What's the integral of \(1\over u\)?

OpenStudy (anonymous):

oh, sorry. Actually it should be ln |u|?

OpenStudy (kinggeorge):

yup. But since \(e^x+8>0\), you can drop the absolute value once you substitute back in for x.

OpenStudy (anonymous):

got it! I guess i get confused as to when I need to stop taking the derivative.. lol

OpenStudy (kinggeorge):

nice job.

OpenStudy (anonymous):

thank you!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!