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OpenStudy (anonymous):
\[\int\limits_{}^{} \frac{ x+1 }{ x^2 + 1}\]
This one?
OpenStudy (anonymous):
yes!
OpenStudy (anonymous):
Separate it into two, try it from here.
\[\int\limits_{}^{} \frac{ x }{ x^2+1} + \int\limits_{}^{} \frac{ 1 }{ x^2+1}\]
OpenStudy (anonymous):
i got up to this part and i know what to do for 1/x^2+1 but when i do x/x^2+1 i get confused
OpenStudy (anonymous):
Use substitution
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OpenStudy (anonymous):
did that and i got u=x^2+1 and du=2xdx which is du/2=xdx
OpenStudy (anonymous):
yup!
OpenStudy (anonymous):
does it end up being 1/2 the integral of 1/u du?
OpenStudy (anonymous):
correct
OpenStudy (anonymous):
thank you
i was looking online but yahoo answers was just confusing me
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OpenStudy (anonymous):
np
hartnn (hartnn):
thats the difference between yahoo answers (answering site) and OpenStudy(study site).
we make sure you learn and can do similar problems on your own by interacting.
Good work! @ECE :)
\(Welcome ~to ~OpenStudy!!~~\ddot \smile \) @arrowhead94