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How can I integrate $\dfrac2{x^3-x^2}$? Can someone please give me some hints? Thanks a lot!

Ali
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darkchampionz
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3 Answers3

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HINT

Think partial fractions.

(for more hints, please ask!)

Parth Thakkar
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HINT:

$$\frac2{x^3-x^2}=\frac2{x^2(x-1)}=\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}$$

for what values of $A,B$, and $C$?

Brian M. Scott
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  • @darkchampionz: No, because it’s essential to combine the partial fractions over their lowest common denominator, which is always the original denominator; you’ve combined them over $x^3(x-1)$ instead. I make it $A=B=-2$, $C=2$. – Brian M. Scott Jul 14 '13 at 07:51
  • @darkchampionz: Yes, that looks much better. :-) – Brian M. Scott Jul 14 '13 at 08:02
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$$\int \frac2{x^3-x^2}\, dx =\frac2{x^2(x-1)}= \int \left(\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}\right)\,dx$$

$$Ax(x-1) + B(x-1) + Cx^2 = 2$$

We choose values for $x$ that "zero out" some of the terms, to simplifying the computation of the unknown values $A, B, C$:

$$x = 1 \implies A(1)(1 - 1) + B(1-1) + C(1^2) = 2 \iff 0 + 0 + C = 2\quad \checkmark$$

$$x = 0 \implies B(-1) = 2 \implies B = -2\quad \checkmark$$

$$x = 2, C = 2, B = -2 \implies 2A - 2 + 8 = 2\iff A = -2\quad \checkmark$$

$$ \int \left(\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}\right)\,dx = \int \left(-\frac{2}x -\frac{2}{x^2}+\frac{2}{x-1}\right)\,dx$$ $$ =-2 \int \frac{dx}x \quad -\quad 2\int \frac{dx}{x^2}\quad + \quad 2 \int \frac{dx}{x-1}$$

amWhy
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