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OpenStudy (cherrilyn):
Evaluate the integral by completing the square and using trig substitution. After completing the square I got (x+3)^2-3 but I don't know what to do next
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OpenStudy (yuki):
you have to give us what we are supposed to integrate is it \[\int\limits[(x+3)^2-2]dx\]?
OpenStudy (anonymous):
no, its unlikely to be that ^ lol it will be something with a division. my guess is : \[1/ [ (x+3)^2 -3] \]
OpenStudy (anonymous):
or possibly the above with a sqrt on the denominator, its a standard integral
OpenStudy (cherrilyn):
\[\int\limits_{}^{} dx \div (x ^{2} + 6x + 6 )^{2}\]
OpenStudy (cherrilyn):
oh no... Sorry.. the equation is \[\int\limits_{}^{} dx / (x ^{2} + 6x + 6)^{2} \]
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