# Evaluating $\int_{0}^{2}\frac{1}{(x+2)(x-1)}dx$

$\int_{0}^{2}\frac{1}{(x+2)(x-1)}$

Will I be able to use partial fractions on the indefinite integral and then evaluate my answer as f(2)-f(0)?

Any help is appreciated as always. :)

EDIT: I looked this up and apparently the integral doesn't converge so now I need help finding where I went wrong.

I had solved for 1 = $\frac{A}{x+2} + \frac{B}{x-1}$

sub in x = 1 and $B=\frac{1}{3}$

sub in x=-2 and $A=\frac{-1}{3}$

Giving:

$\frac{-1}{3}\int\frac{dx}{x+2} + \frac{1}{3}\int\frac{dx}{x-1}$

Simple u-substitution gives:

$\frac{-1}{3}ln|x+2|$ + $\frac{1}{3}ln|x-1| + C$

Which I was going to then evaluate f(2)-f(0) but somewhere I went wrong.

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Given that you had the thought of partial fractions (a good one) you should try it and see what happens. – Ross Millikan Mar 2 '11 at 5:16
$\ln(x-1)$ is unbounded in the range of integration, which violates one of the assumptions of en.wikipedia.org/wiki/Fundamental_theorem_of_calculus (that of continuity). – Ross Millikan Mar 2 '11 at 5:26
But I figured that it didn't matter since they are absolute values. Instead should I split up the result into two parts of a limit? Or should I do something else instead? – Finzz Mar 2 '11 at 5:30
Yes, you should split up the parts and take a limit. The point is that if you try to take the integral up to 1, then down from 1, they both go to to infinity. So you have $\infty - \infty$ which is not well defined. If you take a symmetric limit, you get the en.wikipedia.org/wiki/Cauchy_principal_value but to have a well-defined answer takes more. – Ross Millikan Mar 2 '11 at 5:48
EDIT: {Answered my own question} – Finzz Mar 2 '11 at 6:15

Hint 1: $$\frac{1}{(x+2)(x-1)}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{1}{x+2}\right)$$
Hint 2: Does this converge or diverge? Think about $$\int_{-1}^{1}\frac{1}{x}dx.$$ Does the value of this integral even make sense?
@Finzz: From the definition of integration, (whichever one), the function $\frac{1}{x}$ doesn't behave nicely when we include an interval around zero. You cannot take $\infty$ and subtract $\infty$ to get zero. However, in some sense if we draw the function, it looks like the area under negative part should cancel with the positive part to get 0. In some sense we might think $\int_{-1}^{1}\frac{1}{x}dx=0$. This is a different notion, and is called the "Cauchy Principle Value" en.wikipedia.org/wiki/Cauchy_principal_value , but stick to "it diverges" for your calculus course. – Eric Naslund Mar 2 '11 at 5:44