Integration by parts Q(x) contains irreducible quadratics

[MrGinsu]

Greetings All,

I'm currently trying to integrate this lovely gem.

\(\displaystyle \begin{eqnarray*}
\int f(x) = \int \frac{x+4}{x^2+2x+5}dx
\end{eqnarray*}\)

Now I know I've got to break this into parts using \(\displaystyle \frac{Ax+B}{ax^2+bx+c}\) but this doesn't match the \(\displaystyle x(x^2+1)\) denominator examples I've got in my notes or the problems I've figured out so far.

Here's what I've worked so far:

Break up the integral
\(\displaystyle \int f(x) = \int \frac{x}{x^2+2x+5}dx + \int \frac{4}{x^2+2x+5}dx\)

Now I'm going to work the \(\displaystyle \int \frac{4}{x^2+2x+5}dx\) by parts.

\(\displaystyle \begin{eqnarray*}
\frac{4}{x^2+2x+5} &=& \frac{Ax+B}{x^2+2x+5} \\
4 &=& Ax(x^2+2x+5) + B(x^2+2x+5) \\
4 &=& Ax^3+2Ax +5Ax +Bx^2 +2Bx +5B \\
\end{eqnarray*}\)

I'm now going to get my B value by setting x = 0

if x = 0 then
\(\displaystyle \begin{eqnarray*}
4 &=& 5B \\
\frac{4}{5} &=& B
\end{eqnarray*}\)

Now I'm stuck. How am I going to find A? I don't see how I can get the B part out of the picture by finding an x value the makes it zero.

Any pointers are greatly appreciated.

Thanks in advance!

 

[pka]

Write the expression as: \(\displaystyle \L
\frac{{x + 4}}{{x^2 + 2x + 5}} = \frac{{x + 1}}{{x^2 + 2x + 5}} + \frac{3}{{\left( {x + 1} \right)^2 + 4}}\)

The first fraction is in ln form; the second is in arctan form.

 

[MrGinsu]

Write the expression as: \(\displaystyle \L
\frac{{x + 4}}{{x^2 + 2x + 5}} = \frac{{x + 1}}{{x^2 + 2x + 5}} + \frac{3}{{\left( {x + 1} \right)^2 + 4}}\)

The first fraction is in ln form; the second is in arctan form.

Very cool! Now it is in forms I'm more used to working with. I been staring at this thing for hours. Thanks for the steering, pka.