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Evaluating: $\int (x^{2}+x)\ln|x-1|dx$ My problem is due to the presence of the absolute value.

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As you may think, I think we should break the function into a piecewise function. Of course the function is not differentiable at $x=1$ – Babak S. Jun 13 '12 at 16:11
the derivative of $\ln|x - 1|$ is actually ${1 \over x - 1}$ for all $x \neq 1$, so you can integrate by parts normally when finding the indefinite integral. – Zarrax Jun 13 '12 at 19:30
up vote 2 down vote accepted

I would go along with the suggestion made by Americo Tavares. I'll start where he left off on the first example and finish off the integral.

$$\int \left(2+3u+u^{2}\right) \ln u \ \ du$$

Multiplying through by $\ln u$ leaves us

$$\int \left(u^2 \ln (u) + 3u\ln (u) + 2 \ln (u)\right) \ \ du$$

Splitting the integrand into three separate integrals:

$$\int u^2 \ln (u) \ \ du + 2 \int \ln (u) \ \ du + 3\int u\ln (u) \ \ du$$

Recall integration by parts formula: $\int u \ dv = uv - \int v \ du$

For our first integral $\int u^2 \ln (u) \ \ du$, integrate by parts (since we are using $u$ already, I'll use $r$ and $s$.) Let $r = \ln u, dr = \frac{1}{u} \ \ du, s = \frac{u^3}{3}, ds = u^2 du.$

$$\int u^2 \ln (u) \ \ du + 2 \int \ln (u) \ \ du + 3\int u\ln (u) \ \ du$$

For $\int u \ln u$, let $f$ and $g$ denote our usual $u$ and $v$ in the integration by parts formula above. Use $f = \ln(u), df = \frac{1}{u} \ \ du, dg = u \ \ du, g = \frac{u^2}{2}$

Finally, for $\int \ln u \ \ du$, let $a$ and $b$ represent our usual $u$ and $v.$ Let $a = \ln u, da = \frac{1}{u} \ \ du, db = du, b = u$

After you carry out all of the integration by parts, you should get:

$$- \frac{u^3}{9} + \frac{1}{3} u^3 \ln (u) - \frac{3u^2}{4} + \frac {3}{2}u^2\ln (u) - 2u + 2u\ln(u) + C$$

If you need me to fill in some more details, let me know and I'll be glad to. Note that you may be able to simplify that last expression a bit if it pleases you. I'll leave the second case to you where $x < 1$. Simply comment if you need help with the second example and I'll help.

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Suggestion. Consider two cases:

  1. $x-1>0\Leftrightarrow x>1$ and the substitution $u=x-1$ $$\begin{eqnarray*} \int \left( x^{2}+x\right) \ln \left\vert x-1\right\vert dx &=&\int \left( x^{2}+x\right) \ln \left( x-1\right) dx \\ &=&\int \left( 2+3u+u^{2}\right) \ln u\,du \\ &=&\ldots \end{eqnarray*}$$
  2. $x-1<0\Leftrightarrow x<1$ and the substitution $t=1-x$ $$\begin{eqnarray*} \int \left( x^{2}+x\right) \ln \left\vert x-1\right\vert dx &=&\int \left( x^{2}+x\right) \ln \left( 1-x\right) dx \\ &=&\int \left( -2+3t-t^{2}\right) \ln t\,dt \\ &=&\cdots \end{eqnarray*}$$

Then integrate by parts $$\begin{eqnarray*} \int \ln u\,du &=&u\ln u-u+C \\ \int u\ln u\,du &=&\frac{1}{2}u^{2}\ln u -\int \frac{1}{2}u\,du \\ \int u^{2}\ln u\,du &=&\frac{1}{3}u^{3}\ln u-\int \frac{1}{3}u^{2}\,du \end{eqnarray*}$$

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