# Confused with Integration Problem

• April 24th 2013, 09:43 AM
jdesak
Confused with Integration Problem
Can you please show me how to solve this integral? I = (x+2)/(x^2+x) dx
• April 24th 2013, 09:47 AM
MarkFL
Re: Confused with Integration Problem
Have you tried partial fraction decomposition?
• April 25th 2013, 03:23 AM
x3bnm
Re: Confused with Integration Problem
By partial fraction expansion:

\begin{align*}\frac{A}{x} + \frac{B}{x+1} =& \frac{x+2}{x^2+x}......\text{(1)}\\ A(x+1) + Bx =& x + 2\\ (A+B)x + A =& x + 2\end{align*}

$A + B = 1......\text{(2)}$

$A = 2.............\text{(3)}$

$\therefore\,\,B = -1$

Now plugging these values of $A$ and $B$ into (1) and integrating both sides:

\begin{align*}\int \frac{x+2}{x^2+x}\,\,dx =& \int \left(\frac{A}{x} + \frac{B}{x+1}\right)\,\,dx\\ =& \int \frac{2}{x}\,\,dx - \int \frac{1}{x+1}\,\,dx\\ =& 2\ln{(x)} - \ln{(x+1)} + C\end{align*}
• April 25th 2013, 04:40 AM
Prove It
Re: Confused with Integration Problem
Quote:

Originally Posted by x3bnm
By partial fraction expansion:

\begin{align*}\frac{A}{x} + \frac{B}{x+1} =& \frac{x+2}{x^2+x}......\text{(1)}\\ A(x+1) + Bx =& x + 2\\ (A+B)x + A =& x + 2\end{align*}

$A + B = 1......\text{(2)}$

$A = 2.............\text{(3)}$

$\therefore\,\,B = -1$

Now plugging these values of $A$ and $B$ into (1) and integrating both sides:

\begin{align*}\int \frac{x+2}{x^2+x}\,\,dx =& \int \left(\frac{A}{x} + \frac{B}{x+1}\right)\,\,dx\\ =& \int \frac{2}{x}\,\,dx - \int \frac{1}{x+1}\,\,dx\\ =& 2\ln{(x)} - \ln{(x+1)} + C\end{align*}

Actually it's \displaystyle \begin{align*} 2\ln{|x|} - \ln{|x+1|} + C \end{align*}.