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Integral of x/(x^2+1) from 3 to 1?
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\[\int\limits_{1}^{3}\frac{ x }{ x^2+1 }dx\] ?
i set u = x^2 +1 and the answer i got was 1/2 ln 10 but the answer the book showed was 1/2 ln 5 can someone please explain?
\[\int\limits_{3}^{1}\frac{ x }{ x ^{2}+1 }=\]
u=x^2+1 so du=2x
yeah
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\[\frac{ 1 }{ 2 }*\int\limits_{3}^{1}\frac{ 1 }{ u }du\]
that's ln(u)
1/2*ln(x^2+1) from 3 to 1
1/2ln(10)-1/2ln(2)
1/2 ln u 1/2 ln (x^2+1) from 1 to 3 which would just be 1/2 ln 10 right? the limits are from 1 to 3, not 3 to 1
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so, it is 1/2*ln(10/2)=1/2*ln(5)
ohhh wow. i added wrong.
thanks anyways :)
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