I cant figure out how to integrate this one.
integral (x+2)/((x+3)^2 - 4 ) dx
The ans is
1/4ln(x+5) +1/12 ln(x+1) +c
You can solve this using partial fractions.
$\displaystyle \frac{x + 2}{(x + 3)^2 - 4} = \frac{x + 2}{(x + 3 - 2)(x + 3 + 2)}$
$\displaystyle = \frac{x + 2}{(x + 1)(x + 5)}$.
Applying partial fractions:
$\displaystyle \frac{A}{x + 1} + \frac{B}{x + 5} = \frac{x + 2}{(x + 1)(x + 5)}$
$\displaystyle \frac{A(x + 5) + B(x + 1)}{(x + 1)(x + 5)} = \frac{x + 2}{(x + 1)(x + 5)}$
$\displaystyle A(x + 5) + B(x + 1) = x + 2$
$\displaystyle Ax + 5A + Bx + B = x + 2$
$\displaystyle (A + B)x + 5A+B = 1x + 2$.
Therefore $\displaystyle A+B= 1$ and $\displaystyle 5A+B = 2$.
Solving these simultaneously gives $\displaystyle A = \frac{1}{4}$ and $\displaystyle B = \frac{3}{4}$.
Therefore
$\displaystyle \frac{x + 2}{(x + 3)^2 - 4} = \frac{1}{4(x + 1)} + \frac{3}{4(x + 5)}$.
So $\displaystyle \int{\frac{x + 2}{(x + 3)^2 - 4}\,dx} = \int{\frac{1}{4(x + 1)} + \frac{3}{4(x + 5)}\,dx}$
$\displaystyle = \frac{1}{4}\int{\frac{1}{x + 1}\,dx} + \frac{3}{4}\int{\frac{1}{x + 5}\,dx}$
$\displaystyle = \frac{1}{4}\ln{|x + 1|} + \frac{3}{4}\ln{|x + 5|} + C$.