# Thread: Can someone explain to me what is going on in this integration problem?

1. ## Can someone explain to me what is going on in this integration problem?

This problem has already been worked for me but I am confused about a couple of things. First of all I don't really understand where the -1 + 1 comes from in the first step, and I'm also not really sure why the dx was moved to the numerator in Step 2. Can anyone explain to me what is going on in the first 2 steps? Thanks.

Problem: ∫ (x^2)/(x-1) dx

Step 1:∫ (x^2 -1 + 1)/(x-1) dx

Step 2: ∫ (X^2 -1)/(X-1) dx + (dx)/(x-1)

Step 3:∫ (x-1)(x+1)/(x-1) dx + ∫ (dx)/(x-1)

Step 4:∫ (x+1) dx + ∫ (dx)/(x-1)

Step 5:∫ x dx + ∫ dx + ∫ (dx)/(x-1)

Step 6: [(1/2x)^2] + x + ln [x-1] + C

2. That's a trick commonly used to avoid polynomial long division. Observe that you have an $x-1,$ and if we have $x^2,$ then by addin' & subtractin' 1, we can make a $(x+1)(x-1),$ so we can split the original fraction into two ratios, now

\begin{aligned}
\int{\frac{x^{2}}{x-1}\,dx}&=\int{\frac{\left( x^{2}-1 \right)+1}{x-1}\,dx} \\
& =\int{\frac{(x+1)(x-1)+1}{x-1}\,dx} \\
& =\int{(x+1)\,dx}+\int{\frac{dx}{x-1}}.
\end{aligned}

Does that make sense?

3. Originally Posted by Matt164
This problem has already been worked for me but I am confused about a couple of things. First of all I don't really understand where the -1 + 1 comes from in the first step, and I'm also not really sure why the dx was moved to the numerator in Step 2. Can anyone explain to me what is going on in the first 2 steps? Thanks.

Problem: ∫ (x^2)/(x-1) dx

Step 1:∫ (x^2 -1 + 1)/(x-1) dx

Step 2: ∫ (X^2 -1)/(X-1) dx + (dx)/(x-1)

Step 3:∫ (x-1)(x+1)/(x-1) dx + ∫ (dx)/(x-1)

Step 4:∫ (x+1) dx + ∫ (dx)/(x-1)

Step 5:∫ x dx + ∫ dx + ∫ (dx)/(x-1)

Step 6: [(1/2x)^2] + x + ln [x-1] + C
See steps:

$\int{\frac{x^{2}}{x-1}\,dx}$

$=\int{\frac{\left( x^{2}-1 \right)+1}{x-1}\,dx} \\$ (we just subtract and add 1 to further simplify it)

$=\int{\frac{(x+1)(x-1)+1}{x-1}\,dx} \\$

$=\int{\left( \frac{(x+1)(x-1)}{x-1} + \frac {1}{x-1} \right) \,dx} \\$

$=\int{(x+1)\,dx}+\int{\frac{dx}{x-1}}.$

$= \frac {x^2}{2} +x + \ln (x-1) +c$