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The problem I have is:

Calculate the integral (here $a < -1$)

$$\int_a^{-1} \left( t+\frac{1}{t}\right)\, dt$$

What I tried:

  1. Looked up examples, but couldn't find any where one of the limits of integration was given as an interval like it is in this problem. I might not be using the correct terminology, so am stuck at the start.
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  • $\begingroup$ I am not sure if I follow. You integrate it and you plug in $a$ to the anti derivative? Where is the interval? $\endgroup$ – Sorfosh Jan 21 '18 at 23:39
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    $\begingroup$ $a$ is not an interval, it's just a variable with a value. All they're doing is telling you that its value is less than -1. $\endgroup$ – John Barber Jan 21 '18 at 23:39
  • $\begingroup$ I just mean where the lower limit of integration isn't an exact number. $a<-1$ in this case. $\endgroup$ – LovesPeanutButter Jan 21 '18 at 23:55
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First we find the anti-derivative: $$ \int\left(t + \frac{1}{t}\right)dt = \int t dt + \int \frac{dt}{t} = \frac{t^2}{2} + \ln |t| + c. $$ Thus, for $c<d\le -1$, $$ \int_c^d \left(t + \frac{1}{t}\right)dt = \left[ \frac{t^2}{2} + \ln |t| \right]_c^d = \left[ \frac{d^2}{2} + \ln |d| - \frac{c^2}{2} - \ln |c|\right] = \frac{d^2-c^2}{2} + \ln \left|\frac{d}{c}\right|. $$

Can you apply this to your specific problem?

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  • $\begingroup$ I follow the first step, but am having trouble understanding the second step. Why is the interval $c<d\le -1$? $\endgroup$ – LovesPeanutButter Jan 21 '18 at 23:52
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    $\begingroup$ @LovesLinearAlgebra It is a generalization of your integral. In your case, $c=a$ and $d=-1$. $\endgroup$ – dromastyx Jan 22 '18 at 0:13
  • $\begingroup$ @LovesLinearAlgebra this may not be true if $c < 0 < d$ for example because at $t=0$ the integrand is problematic.. $\endgroup$ – gt6989b Jan 22 '18 at 1:26
  • $\begingroup$ But arent we always below $-1$ at the top limit of integration is $-1$ and $a <-1$? So we dont have to worry about evaluating at $0$? $\endgroup$ – LovesPeanutButter Jan 22 '18 at 1:56

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