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Evaluate $$I=\int \frac{dx}{x(x-1)^3(x-2)^2}$$ without using tedious partial fractions.

My Try:

we have $$1=\left((x-1)-(x-2)\right)^2$$ So $$=\int \frac{\left((x-1)-(x-2)\right)^2dx}{x(x-1)^3(x-2)^2}$$ So

$$I=\int \frac{dx}{x(x-1)(x-2)^2}+\int \frac{dx}{x(x-1)^3}-2\int \frac{dx}{x(x-1)^2(x-2)} $$

any clue here?

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  • $\begingroup$ You will have to do partial fractions looks like $\endgroup$ – K Split X Sep 10 '17 at 18:12
  • $\begingroup$ Also the integration is split within many terms, and so it seems evident that it is first integrated via partial fractions, and then to its respective antiderivatives $\endgroup$ – K Split X Sep 10 '17 at 18:17
  • $\begingroup$ By "without using tedious partial fractions", do you mean that you can use partial fractions only if its application is easy, or that you cannot use it at all? $\endgroup$ – projectilemotion Sep 10 '17 at 18:34
  • $\begingroup$ If you find tedious doing partial fractions (and I agree with you) you can always make others do it for you and pretend to have done them by yourself :) symbolab.com/solver/partial-fractions-calculator/… $\endgroup$ – Raffaele Sep 10 '17 at 20:16
  • $\begingroup$ If someone thinks that using web site to do partial fraction is cheating, my opinion is that also use calculator to to $23.708^{\frac{e}{\pi}}$ is cheating :) $\endgroup$ – Raffaele Sep 10 '17 at 20:19
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You can reduce the labour a little by writing

$\dfrac{1}{x(x-1)^3(x-2)^2} = \dfrac{(1-x+x^2)+x(x-1)}{x(x-1)^3(x-2)^2}$

$=\dfrac{x(x-2)^2-(x-1)^3}{x(x-1)^3(x-2)^2} + \dfrac{1}{(x-1)^2(x-2)^2}$

$=\dfrac{1}{(x-1)^3} - \dfrac{1}{x(x-2)^2}+\dfrac{1}{(x-1)^2(x-2)^2}$

$=\dfrac{1}{(x-1)^3}- \dfrac{1}{x(x-2)^2}+ \dfrac{1}{(x-1)^2}+\dfrac{1}{(x-2)^2}+2 \left[\dfrac{1}{x-1} - \dfrac{1}{x-2} \right]$

If you wish you can further decompose $\dfrac{1}{x(x-2)^2}= \dfrac{1}{(x-2)^2}+\dfrac{1}{2} \left[\dfrac{1}{x} - \dfrac{1}{x-2} \right]$

This expression can be readily integrated.

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