# Integrating $x/(x-2)$ from $0$ to $5$

How would one go about integrating the following?

$$\int_0^5 \frac{x}{x-2} dx$$

It seems like you need to use long division, split it up into two integrals, and the use limits. I'm not quite sure about the limits part.

• It will turn out that the integral does not exist. That may be clearer if you drag the bad point to $0$, by letting $x-2=u$. – André Nicolas Mar 22 '13 at 19:43

The integrand $\frac{x}{x-2}$ has a vertical asymptote at $x=2$. Therefore the integral $$\begin{equation*} I=\int_{0}^{5}\frac{x}{x-2}dx=\int_{0}^{2}\frac{x}{x-2}dx+\int_{2}^{5}\frac{x }{x-2}dx=I_{1}+I_{2} \end{equation*}$$ is an improper integral of the second kind. It exists only if both integrals $I_{1}$ and $I_{2}$ exist. Let's compute the first one using the substitution $u=x-2,dx=du$, as commented above $$\begin{eqnarray*} I_{1} &=&\int_{0}^{2}\frac{x}{x-2}dx=\int_{-2}^{0}\frac{u+2}{u} du=\int_{-2}^{0}1du+\int_{-2}^{0}\frac{2}{u}du \\ &=&2+2\int_{-2}^{0}\frac{1}{u}du=2+2\ln \left\vert u\right\vert _{-2}^{0}=2+2\lim_{\varepsilon \rightarrow 0}\ln \left\vert u\right\vert _{-2}^{-\varepsilon },\quad \varepsilon >0 \\ &=&2+2\lim_{\varepsilon \rightarrow 0}\ln \left\vert -\varepsilon \right\vert -2\ln \left\vert -2\right\vert \\ &=&2-2\ln 2+2\lim_{\varepsilon \rightarrow 0}\ln \varepsilon . \end{eqnarray*}$$ Since $\lim_{\varepsilon \rightarrow 0}\ln \varepsilon$ ($\varepsilon>0$) doesn't exist, the integral $\int_{-2}^{0}\frac{2}{u}du$ does not converge and so does $I_{1}$. Similarly we might conclude that $I_{2}$ does not converge as well.

Yes, exactly, you do want to use "long division"...

Note, dividing the numerator by the denominator gives you:

$$\int_0^5 {x\over{x-2}} \mathrm{d}x = \int_0^5 \left(1 + \frac 2{x-2}\right) \mathrm{d}x$$

Now simply split the integral into the sum of two integrals:

$$\int_0^5 \left(1 + \frac 2{x-2}\right) \mathrm{d}x \quad= \quad\int_0^5 \,\mathrm{d}x \;\; + \;\; 2\int_0^5 \frac 1{x-2} \,\mathrm{d}x$$

The problem, of course, is what happens with the limits of integration in the second integral: If $u = x-2$ then the limits of integration become $\big|_{-2}^3$, and we will see that the integral does not converge - there is a discontinuity at $u = 0$, and indeed, a vertical asymptote at $x = -2$, hence the limit of the integral - evaluated as $u \to 0$ does not exist, so the integral does not converge. And so the sum of the integrals will not converge.

Shashwat, the antiderivative of $1/x$ is not $\ln(x)$, but $\ln|x|$. So in the computation there wouldn't be any natural log of $-2$. The reason that this integral does not exist, is as stated earlier, the discontinuity issues at $x=2$.

$\displaystyle \int$$\displaystyle {\frac{{dx}}{{ax + b}}} = \displaystyle {\frac{{1}}{{a}}}$$\displaystyle \int$$\displaystyle {\frac{{dt}}{{t}}}$ = $\displaystyle {\frac{{1}}{{a}}}$ $\ln| t| + C =$ $\displaystyle {\frac{{1}}{{a}}}\ln| ax + b| + C$.

where

$ax + b = t$

• You might also want to add a note about what happens with those limits of integration. – Amzoti Mar 22 '13 at 20:14
• And the $x$ on the numerator. – Daryl Mar 22 '13 at 20:19