How to integrate $\frac{x+4}{x^2+2x+5}$

Question

Having a little trouble on how to break it up. How to integrate $$\frac{x+4}{x^2+2x+5}$$

Answer 1

Since $x^2+2x+5 = (x+1)^2+2^2$, we should split it up as

$\dfrac{x+4}{x^2+2x+5} = \dfrac{x+1+3}{x^2+2x+5} = \dfrac{x+1}{(x+1)^2+2^2}+\dfrac{3}{(x+1)^2+2^2}$.

Both of these should be easy to integrate.

Answer 2

You may want to try this approach:

• Let $r_{i}$ be the complex roots of the polynomial $x^2 + 2x+5$, $r_{1,2} = -1 \pm 2 \mathrm{i}$.
• Perform a partial fraction decomposition of the integrand: $$Q(x) = \frac{x+4}{x^2+ 2x +5} = \frac{A}{x-r_1} + \frac{B}{x-r_2},$$

• Then the coefficients $A$ and $B$ can be computed as follows: $$\begin{align} &A = \lim_{x \to r_1} (x-r_1) Q(x) = \frac{r_1 +4}{r_1 -r_2}, \\ &B = \lim_{x \to r_2} (x-r_2) Q(x) = \frac{r_2 +4}{r_2 -r_1}, \end{align}$$ and $A = \overline{B}$ (complex conjugates).

• Integrate and simplify: $$I = \int Q(x) \, \mathrm{d}x = A \log{|x-r_1|} + B \log{|x-r_2|}.$$ The result should remain real.

Cheers!