1. ## Integration

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

2. It it $\displaystyle \int{\frac{x+2}{(x+3)^2 - 4}\,dx}$ or $\displaystyle \int{\frac{x+2}{(x+3)^2}-4\,dx}$?

3. It the 1st one .
thanks prove it

\int{\frac{x+2}{(x+3)^2 - 4}\,dx}

4. 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 $\displaystyle \int{\frac{x + 2}{(x + 3)^2 - 4}\,dx} = \int{\frac{1}{4(x + 1)} + \frac{3}{4(x + 5)}\,dx}$

$\displaystyle = \frac{1}{4}\int{\frac{1}{x + 1}\,dx} + \frac{3}{4}\int{\frac{1}{x + 5}\,dx}$

$\displaystyle = \frac{1}{4}\ln{|x + 1|} + \frac{3}{4}\ln{|x + 5|} + C$.

5. Originally Posted by Prove It
You can solve this using partial fractions.

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

$\displaystyle = \frac{x + 3}{(x + 1)(x + 5)}$.

...
Very interesting the 'identity' $\displaystyle \frac{x + 2}{(x + 3 - 2)(x + 3 + 2)}= \frac{x + 3}{(x + 1)(x + 5)}$...from which it derives that $\displaystyle x+2=x+3$...

Kind regards

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

6. I do hate when typos go all the way through the post... Fixed now...