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I am trying to compute the integral

$$ \int\frac{\ln(x)}{(x-3)^2}\mathrm{d}x $$

I have tried the following substitution, but seem to get nowhere: $u = x - 3$.

$$x=u+3$$ $$dx=du$$ $$\int\frac{\ln(u+3)}{u^3}\mathrm{d}u$$

I get stuck at this point and do not know what to do, could anyone lead me in the right direction?

David K
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JoshBoiskin
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2 Answers2

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Note that $u$ is squared, not cubed, in the denominator.

$$I = \int\frac{\ln(u+3)}{u^3}\mathrm{d}u$$

Using integration by parts, take $$w = \ln(u+3) \implies dw= \frac{1}{u+3}\,du$$ Then $$dv = \frac 1{u^2} \implies v = -\frac{1}{u}$$

$$I = wv - \int v\,dw$$

$$I = -\frac{\ln(u + 3)}{u} + \int \frac 1{u}\cdot \frac 1{u+3}\,du$$

Now you can use partial fraction decomposition $$\int \frac 1{u(u+3)}\,du =\int\left(\frac Au + \frac B{u+1}\right)\,du$$

Or, you can complete the square in the denominator of the remaining integral to use trig substitution. $$u(u+3) = u^2 + 3u + \frac 94 -\frac 94 = \left(u + \frac 32\right)^2 - \left(\frac 32\right)^2$$

Now put $\sec \theta = \frac 32\left(u + \frac 32\right)$, and use the fact that $$\sec^2 \theta -1 = \tan^2\theta$$

amWhy
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Let $u=\ln x$ and $dv=\frac{1}{(x-3)^2} dx$, so $du=\frac{1}{x} dx$ and $v=-\frac{1}{x-3}$.

Then $$\displaystyle\int\frac{\ln x}{(x-3)^2} \, dx=uv-\int v\,du=-\frac{\ln x}{x-3}-\int-\frac{1}{x(x-3)}\, dx=-\frac{\ln x}{x-3}+\int\frac{1}{x(x-3)}\, dx.$$

Now use partial fractions:

$$\frac{1}{x(x-3)}=\frac{A}{x}+\frac{B}{x-3}\implies A(x-3)+Bx=1\implies A=-\frac{1}{3} \text{ and } B=\frac{1}{3}.$$

Michael Hardy
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user84413
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  • Why do you use \displaystyle when actually displaying the line would clearly work better? I've taken the liberty of changing two lines to dipslayed lines. Lots of posters here do that; I've never known why. ${}\qquad{}$ – Michael Hardy Aug 26 '14 at 17:51
  • Thanks for making this change -- it's mainly due to my lack of familiarity with Latex. – user84413 Aug 26 '14 at 17:56