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integrate 1/((x-3)^2 +1)
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i sorta know how to do it but I'm not sure why theres a negative in front of the tan^-1
\[\int\frac{1}{(x-3)^2+1} dx\] ?
yeah thats it
u sub is the way to go u= x-3 du= dx \[\int\frac{1}{u^2+1} du\]
yep got that far but I'm missing a negative sign infront of my answer lol
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we know this is equal to \[arctan(u)+C\]
u=x-3 hence our answer is \[arctan(x-3)+C\]
there is a negative sign in front of it in the answers
\[-arctan(3-x)=arctan(x-3)\]
oh yeah lol thats right my bad i ended up with the right answer. thanks:)
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