# Thread: Help on an indefinte integral

1. ## Help on an indefinte integral

$\displaystyle\int\frac{x^2+2x}{(x+1)^2}dx$

Can someone help me with a starting point on this one please? Thanks

2. Is...

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

... so that...

Kind regards

$\chi$ $\sigma$

3. Can you show me how those are equal?

4. Originally Posted by dbakeg00
Can you show me how those are equal?
You should be able to do the algebra at this level.

$\displaystyle \frac{x^2 + 2x}{x^2 + 2x + 1} = \frac{(x^2 + 2x + 1) - 1}{x^2 + 2x + 1} = ....$

5. Originally Posted by chisigma
Is...

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

... so that...

Kind regards

$\chi$ $\sigma$
$\displaystyle\int 1-\frac{1}{(x+1)^2}dx$

$\displaystyle\int 1\ dx - \int (x+1)^{-2}dx$

$\displaystyle x+(x+1)^{-1}+c$

6. Well done

7. The book is giving me an answer of:

$\displaystyle\frac{x^2}{(x+1)}+c$

8. $\begin{array}{cl}x+(x+1)^{-1}+c &= x + \dfrac{1}{x+1} + c \\

& \\

&= \dfrac{x(x+1) + 1}{x+1} + c \\ & \\

& = \dfrac{x^2 + x + 1}{x+1} + c \end{array}$

There must be a mistake in the book.

EDIT: It can be broken into two fractions:

$\dfrac{x^2 + x + 1}{x+1} + c = \dfrac{x^2}{x+1} + \dfrac{x+1}{x+1} + c = \dfrac{x^2}{x+1} +1 + c = \dfrac{x^2}{x+1} +c'$

9. Ok. I came up with the same thing when I expanded it. Just wanted to make sure I wasn't overlooking something. Thanks again for the help!

10. Originally Posted by dbakeg00
The book is giving me an answer of:

$\displaystyle\frac{x^2}{(x+1)}+c$
$\displaystyle \frac{x^2}{(x+1)}+c$ is equivalent to $\displaystyle x + \frac{1}{(x+1)} + d$:

$\displaystyle x + \frac{1}{(x+1)} + d = \frac{x^2 + x + 1}{x + 1} + d = \frac{x^2 + (x + 1)}{x + 1} + d = \frac{x^2}{x + 1} + 1 + d = \frac{x^2}{x + 1} + c$.