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

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.

  • 0
    Thanks, it's that I forgot that you cannot take the natural log of 0.2011-03-02

1 Answers 1

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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?

  • 4
    @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" http://en.wikipedia.org/wiki/Cauchy_principal_value , but stick to "it diverges" for your calculus course.2011-03-02