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I can get started in the right direction, but cant seem to get all of the way there, and any examples I can find don't have the same complications.

$$\int {2x\over (x-1)^2}\cdot dx$$

What I have tried:

I can't integrate this with the power rule, so my next step was to use ln and u-substitution, since I don't see how long division could help. I also recognized that I could factor out a 2.

I expanded the denominator: $(x-1)^2 = (x-1)\cdot(x-1) = x^2 -2x +2$

Let $u$ = $x^2 - 2x + 2 \therefore du = 2x - 2 \cdot dx$ using the power rule. , leaving me with:

$$2 \cdot \int {x \over u}\cdot dx$$

The problem is, $du = 2x-2 \cdot dx \neq x \cdot dx$. Even if I didn't factor out the 2 in the beginning I dont see how I could rectify this.

I don't see any way to get into the form of $\int {u'\over u} \cdot dx$ which I need to do.

I checked the answer in the book, which is $2\ln(\lvert x-1 \rvert) - {2 \over x-1} + C $

From that, I still cannot figure it out, but I do recognize that $2 \ln (|x-1|) = \ln((x-1)^2)$ as in the original denominator.

I'm stuck.

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  • $\begingroup$ Do you know what partial fractions are? $\endgroup$ – Bob Krueger Apr 11 '15 at 16:33
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    $\begingroup$ I would try setting u to be x-1. A simple denominator results in a simpler fraction. $\endgroup$ – user1337 Apr 11 '15 at 16:33
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    $\begingroup$ $(x - 1)^{2} = x^{2} - 2x + 1$ not $x^{2} - 2x + 2$.. $\endgroup$ – mattos Apr 11 '15 at 16:34
  • $\begingroup$ @Mattos Oh. My God. I'd like to think I'm pretty darn good at math, but this takes the cake for the stupidest mistake I've ever made. $\endgroup$ – Bassinator Apr 11 '15 at 16:35
  • $\begingroup$ @HCBPshenanigans Don't worry, we all make them. And use user1337's hint to solve your integral. $\endgroup$ – mattos Apr 11 '15 at 16:36
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Let $x-1 = u$. Then $dx = du$ and $x=u+1$

This gives you the integral $$\begin{align} \int \frac{2(u+1)}{u^2}\,du &= 2\int \left(\frac 1u + u^{-2}\right)\,du\\ \\ &= 2\left(\ln|u| -u^{-1}\right) + C\\ \\ & = 2\ln|x-1| - \frac 2{u} + C\\ \\ &= \ln(x-1)^2 - \frac 2{x-1} + C \end{align}$$

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  • $\begingroup$ I think I can take it from here, lemme work some numbers and see where I get. $\endgroup$ – Bassinator Apr 11 '15 at 16:38
  • $\begingroup$ Much appreciated! Once I realized that when I let $u = x-1$ that it could be rearranged as $x = u + 1$ it was so much easier. $\endgroup$ – Bassinator Apr 11 '15 at 16:51
  • $\begingroup$ You're welcome! $\endgroup$ – Jordan Glen Apr 11 '15 at 16:54
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Hint: You can rewrite the numerator as $$2x=2x-2+2=2(x-1)+2$$

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  • $\begingroup$ I see the validity, but how does that help (I think I see where you are going but I'd like to hear your words.) and how would I know to do it? $\endgroup$ – Bassinator Apr 11 '15 at 16:37
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    $\begingroup$ Your edit just clarified it a lot. $\endgroup$ – Bassinator Apr 11 '15 at 16:38
  • $\begingroup$ Can you proceed? $\endgroup$ – Fermat Apr 11 '15 at 16:42
  • $\begingroup$ I think so. Working some numbers. $\endgroup$ – Bassinator Apr 11 '15 at 16:45
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$$\int \frac{2x}{(x-1)^2} dx$$ The degree of numerator is 1 and of denominator is 2. Difference +1. This trick would work: $$N^r=A(D^r)'+B$$ Or: $$2x=A((x-1)^2)'+B=2A(x-1)+B$$ You can solve this comparing coefficients of same powers of x or by just putting some random values in there: $$2x=2Ax+B-2A$$ Now: $$2A=A\implies A=1,B-2A=0\implies B=2A=2$$ So: $$\int \frac{2x}{(x-1)^2} dx=\int \frac{[(x-1)^2]'+2}{(x-1)^2} dx=\int \frac{[(x-1)^2]'}{(x-1)^2} dx+\int \frac{2}{(x-1)^2} dx$$ And we know: $$\int \frac{f'}{f}dx=\ln f+c\text{ and }\int\frac{dx}{(ax+b)^2}\stackrel{u=ax+b}=\int\frac1a\frac{du}{u^2}=-\frac1{au}+c=-\frac1{a(ax+b)}+c$$ So: $$\int \frac{2x}{(x-1)^2} dx=\ln((x-1)^2)+2\ln|x-1|+c\\ =2\ln|x-1|-2\frac1{x-1}+c$$

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