1. ## Evaluating Partial Fractions

Hi,
This question is bugging me.

Evaluate the Integral

(x)/(x^2+7x+6) from 0 to 4

Solving for it I get (6/5)ln (x+6) - (1/5) ln (x+1)

Plugging values in I get (6/5) ln 10 - (6/5) ln 6 - (1/5) ln 5

The answer is (-6/5) ln 3 + ln 5.

2. Originally Posted by shogunhd
Evaluate the Integral

(x)/(x^2+7x+6) from 0 to 4
Since I don't use partial fractions technique frequently, you can do this as follows:

$\int_0^4 {\frac{x}
{{x^2 + 7x + 6}}\,dx} = \frac{1}
{5}\int_0^4 {\frac{{6(x + 1) - (x + 6)}}
{{(x + 1)(x + 6)}}\,dx} .$

So $\int_0^4 {\frac{x}
{{x^2 + 7x + 6}}\,dx} = \frac{1}
{5}\left( {6 \cdot \ln \left| {x + 6} \right|\Big|_0^4 - \ln \left| {x + 1} \right|\Big|_0^4} \right).$

Hence $\int_0^4 {\frac{x}
{{x^2 + 7x + 6}}\,dx} = \frac{1}
{5}\left( {6\ln 10 - 6\ln 6 - \ln 5} \right).$

Now here's the make-up:

\begin{aligned}
\frac{1}
{5}\left( {6\ln 10 - 6\ln 6 - \ln 5} \right) &= \frac{1}
{5}\left( {6\ln 2 + 6\ln 5 - 6\ln 2 - 6\ln 3 - \ln 5} \right)\\
&= \frac{1}
{5}\left( {5\ln 5 - 6\ln 3} \right)\\
&= \ln 5 - \frac{6}
{5}\ln 3,\end{aligned}