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DeMath · #6 January 2nd, 2009, 01:05 PM

Quote:

Originally Posted by frog09

$\int_1^3 (x+3)/(x^2+6x) dx = ?$

I tired separating and simplifying the problem into two different integrals:

$\int_1^3 1/(x+6)$ + $\int_1^3 3/(x^2+6x)$

and get...
$\int_1^3 ln(x+6) + \int_1^3 3ln(x^2+6x)$

evaluating I get

$[ln9-ln7]+ 3[ln27+ln7]$

but that is not any of the solutions.

$\begin{gathered}\int\limits_1^3 {\frac{{x + 3}}{{x^2 + 6x}}dx} = \frac{1}{2}\int\limits_1^3 {\frac{{2x + 6}}{{x^2 + 6x}}dx = } \frac{1}{2}\int\limits_1^3 {\frac{{d\left( {x^2 + 6x} \right)}}{{x^2 + 6x}} = } \hfill \\= \left. {\frac{1}{2}\ln \left( {x^2 + 6x} \right)} \right|_1^3 = \frac{1}{2}\left( {\ln 27 - \ln 7} \right) = \frac{1}{2}\left( {3\ln 3 - \ln 7} \right). \hfill \\\hfill \\ \end{gathered}$