# Thread: Integrate [(x^3+x^2-4x)/(x^2-4x+4)] by parts.

1. ## Integrate [(x^3+x^2-4x)/(x^2-4x+4)] by parts.

Hello.

I know I have done something very wrong, but not sure where.
We have just learnt how to integrate by parts.

I have uploaded a scan of my working in this link :
http://imageshack.us/m/842/448/mhf.png
(It gave me an error when I tried to insert image into this post)

Thank you very much .

2. You don't use integration by parts. Here, since the numerator is of greater degree than the denominator, you first have to long divide.

$\displaystyle \displaystyle \frac{x^3 + x^2 - 4x}{x^2 - 4x + 4} = x + 5 + \frac{12x - 20}{x^2 - 4x + 4}$.

Then you need to factorise the denominator of the proper fraction and apply the partial fractions technique.

3. Originally Posted by anees7112
Hello.

I know I have done something very wrong, but not sure where.
We have just learnt how to integrate by parts.

I have uploaded a scan of my working in this link :
http://imageshack.us/m/842/448/mhf.png
(It gave me an error when I tried to insert image into this post)

Thank you very much .
With the substitution $\displaystyle t=x-2$ function to be integrated becomes...

$\displaystyle \frac{x^{3} + x^{2} - 4 x}{x^{2}-4 x +4}= t+7+\frac{12}{t}+ \frac{4}{t^{2}}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Thank you both for the answer. I am able to do the question now