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


*

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

*$x-1<0\Leftrightarrow x<1$ 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
$$\begin{eqnarray*}
\int \ln u\,du &=&u\ln u-u+C \\
\int u\ln u\,du &=&\frac{1}{2}u^{2}\ln u -\int \frac{1}{2}u\,du
\\
\int u^{2}\ln u\,du &=&\frac{1}{3}u^{3}\ln u-\int \frac{1}{3}u^{2}\,du
\end{eqnarray*}$$
A: 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.
