Antiderivative fraction
I'm would like to find the antiderivative of
$$\frac{x^2}{(x-1)^5}$$
I tried without using partial fractions but I did not manage, so I started a lesson on partial fractions but I do not succeed in proving $$\frac{x²}{(x-1)^5}=\frac{1}{(x-1)^3}+\frac{2}{(x-1)^4}+\frac{1}{(x-1)^5}$$
Could you tell me how to proceed? Thank you
$\endgroup$3 Answers
$\begingroup$Do a substitution. Let $u = (x-1)$. This means that $x^2 = (u+1)^2$ and the denominator is $u^5$. Expand the numerator and integrate as usual.
$\endgroup$ 1 $\begingroup$Indeed you are correct that,
$$\frac{x^2}{(x-1)^5} = \frac{1}{(x-1)^3} + \frac{2}{(x-1)^4} + \frac{1}{(x-1)^5}$$
One can integrate each of these terms in turn. I will do the first to help. Let $u=x-1$ and $du=dx$ then,
$$ \int\frac{1}{(x-1)^3} dx = \int u^{-3} du = -\frac{1}{2}u^{-2} = -\frac{1}{2}\frac{1}{(x-1)^2} + c$$ where $c$ is the constant of integration. The final answer is, $$ -\frac{1}{2}\frac{1}{(x-1)^2} -\frac{2}{3}\frac{1}{(x-1)^3} -\frac{1}{4}\frac{1}{(x-1)^4}+c $$
$\endgroup$ $\begingroup$This is a different way of looking at David's solution, but without doing a substitution. You want factors of $(x-1)^r$ in the numerator, so you can cancel them with the denominator.
Looking at the highest term, which is $x^2$, it seems that $(x-1)^2$ is a good place to start, because it takes care of the highest order term.
Write $x^2=(x-1)^2+p(x)$ then $p(x)=x^2-(x-1)^2=2x-1$
Now $p(x)=2(x-1)+1$ so that $x^2=(x-1)^2-2(x-1)+1$ and you get the three terms you are looking for.
This looks like a trick. The method of partial factions (which has its own tricks for efficient computation) automates such tricks in a way which always works. Occasionally there is a simpler way of looking at things, like the substitution here. Only practice will help you to develop efficient methods and notice short cuts.
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