# Math Help - another integration

1. ## another integration

integrate (x-3)/(x^2 + 2x+4).

2. Originally Posted by twilightstr
integrate (x-3)/(x^2 + 2x+4).
note that $\int \frac {x - 3}{x^2 + 2x + 4}~dx = \int \frac {x + 1}{x^2 + 2x + 4}~dx - \int \frac 4{x^2 + 2x + 3}$

for the first integral, a substitution of $u = x^2 + 2x + 4$ works.

for the second integral, note that $\frac 4{x^2 + 2x + 4} = \frac 4{(x + 1)^2 + 3} = \frac 43 \cdot \frac 1{\left( \frac {x + 1}{\sqrt{3}} \right)^2 + 1}$

write the integral in that form and then do a substitution of $u = \frac {x + 1}{\sqrt{3}}$. you should be able to handle it from there

3. Originally Posted by twilightstr
integrate (x-3)/(x^2 + 2x+4).
$\frac{x-3}{x^2+2x+4}=\frac{x-3}{(x+1)^2+3}$

so try the substitution: $u=x+1$ to get:

$\frac{x-3}{x^2+2x+4}=\frac{x-3}{(x+1)^2+3}=\frac{u}{u^2+3}-\frac{4}{u^2+3}$

CB