I keep messing up with the integration part of this I think. Before that I have to factor out the $x$ on the bottom, and then set up the $A$ and $B$ right?
Evaluate the integral. (Remember to use $\ln|u|$ where appropriate. Use $C$ for the constant of integration.) $$\int\frac{7x^2+2x-7}{x^3-x}\,dx$$
On
Hint : Apply http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Procedure
$$ \frac{7x^2+2x-7}{x^3-x} = \frac{7x^2+2x-7}{x(x^2-1)} = \frac{7x^2+2x-7}{x(x-1)(x+1)} $$
$$ \frac{7x^2+2x-7}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{(x-1)} + \frac{C}{(x+1)} $$
You can obtain A,B,C by setting the value of x are 1,-1,0 respectively and compare the coefficient of different power of x i.e 2,1 and 0.
$$ \int (\frac{A}{x} + \frac{B}{(x-1)} + \frac{C}{(x+1)}) dx = \int \frac{A}{x} + \int \frac{B}{(x-1)} + \int \frac{C}{(x+1)}$$
Now, apply this, to get your final answer
$$ \int \frac{1}{x+k} = ln|x+k| + C$$
HINT: As $x^3-x=x(x^2-1)=x(x+1)(x-1),$
Using Partial fraction decomposition
$$\frac{7x^2+2x-7}{x(x-1)(x+1)}=\frac A{x+1}+\frac Bx+\frac C{x-1}$$
Multiply either sides by $x^3-x=x(x^2-1)=x(x+1)(x-1)$
and compare the coefficients of the different power of $x$ to find $A,B,C$
Finally use $\displaystyle \int\frac{dx}{x+m}=\ln|x+m|+C$