Basic integration problem

[jorapello]

Hi, I'm having trouble understanding this problem:

integral of (x^2) / (x-1) dx

Every source says the answer is 1/2x^2 + x + ln|x-1| + C.

I understand 1/2x^2 and ln|x-1| but I don't see where the +x in the middle comes from.

In addition, my book says that the original problem becomes:

integral of (x+1)dx + integral of (1/(x-1))dx

How does the x^2 become x+1? Or am I missing something else?

Thanks,
John

 

[daon]

Add and subtract 1 from the numerator and split it into (x^2-1)/(x-1) + 1/(x-1) = (x-1)(x+1)/(x-1) + 1/(x-1) = x + 1 + 1/(x-1) , then integrate

 

[soroban]

Hello, John!

\(\displaystyle \int \frac{x^2}{x-1}\,dx\)

\(\displaystyle \text{Long division: }\;\frac{x^2}{x-1} \;=\;x + 1 +\frac{1}{x-1}\)


\(\displaystyle \text{Hence: }\; \int\frac{x^2}{x+1}\,dx \;=\;\int\left[x + 1 + \frac{1}{x-1}\right]\,dx \;=\;\int(x + 1)\,dx + \int\frac{dx}{x-1}\)


\(\displaystyle \text{Get it?}\)