tricky integral

[roxstar1]

integral [(x^2)/(x-1)] dx

I am stuck trying to integrate this problem. U-substitution isn't working and I can't (or don't know how to) seperate the integral and solve it that way. Could someone please point me in the right direction?

 

[Matt]

Method #1. u-substitution.

Let \(\displaystyle \L u = x - 1.\) Then \(\displaystyle \L du=dx\) and \(\displaystyle \L x=u+1.\)

\(\displaystyle \L \qquad\begin{eqnarray*}
\int\frac{x^2}{x-1}\,dx
&=&\int\frac{(u+1)^2}{u}\,du\\
&=&\int\frac{u^2+2u+1}{u}\,du\\
&=&\int\left(u+2+\frac1u\right)\,du
\end{eqnarray*}\)

and so on.

Method #2. a "clever" separation.

\(\displaystyle \L \qquad\begin{eqnarray*}
\int\frac{x^2}{x-1}\,dx
&=&\int\frac{x^2-1+1}{x-1}\,du\\
&=&\int\frac{x^2-1}{x-1}\,dx+\int\frac{1}{x-1}\,dx\\
&=&\int\frac{(x-1)(x+1)}{x-1}\,dx+\int\frac{1}{x-1}\,dx\\
&=&\int(x+1)\,dx+\int\frac{1}{x-1}\,dx
\end{eqnarray*}\)

The rest is history...

 

[roxstar1]

didn't think of doing that. thanks a lot