$\displaystyle \displaystyle\int\frac{x^2+2x}{(x+1)^2}dx$
Can someone help me with a starting point on this one please? Thanks
$\displaystyle \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:
$\displaystyle \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' $
$\displaystyle \displaystyle \frac{x^2}{(x+1)}+c$ is equivalent to $\displaystyle \displaystyle x + \frac{1}{(x+1)} + d$:
$\displaystyle \displaystyle x + \frac{1}{(x+1)} + d = \frac{x^2 + x + 1}{x + 1} + d = \frac{x^2 + (x + 1)}{x + 1} + d = \frac{x^2}{x + 1} + 1 + d = \frac{x^2}{x + 1} + c$.