1. Confused with Integration Problem

Can you please show me how to solve this integral? I = (x+2)/(x^2+x) dx

2. Re: Confused with Integration Problem

Have you tried partial fraction decomposition?

3. Re: Confused with Integration Problem

By partial fraction expansion:

\displaystyle \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*}

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

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

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

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

\displaystyle \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*}

4. Re: Confused with Integration Problem

Originally Posted by x3bnm
By partial fraction expansion:

\displaystyle \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*}

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

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

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

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

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