# How do you evaluate int 1/(x+7)(x^2+9) for [-1, 3]?

Sep 16, 2015

$- 24 + 58 \ln \left(\frac{5}{3}\right)$

#### Explanation:

the given $\frac{{x}^{2} + 9}{x + 7}$ can be written as $x + \frac{- 7 x + 9}{x + 7}$
which can be furthur reduced as $x - 7 + \frac{58}{x + 7}$
${\int}_{-} {1}^{3}$$x - 7 + \frac{58}{x + 7} \mathrm{dx}$=${\int}_{-} {1}^{3}$$\frac{{x}^{2} + 9}{x + 7}$$\mathrm{dx}$
${\int}_{-} {1}^{3}$$x \mathrm{dx} - 7 \mathrm{dx} + \frac{58}{x + 7} \mathrm{dx}$
${x}^{2} / 2 - 7 x + 58 \ln \left(x + 7\right)$from $\left[- 1 3\right]$
$\frac{9}{2} - 7 \cdot 3 + 58 \ln 10$-$\left(\frac{1}{2} - 7 \cdot \left(- 1\right) + 58 \ln 6\right)$
$4 - 28 + 58 \ln \left(\frac{10}{6}\right)$
$- 24 + 58 \ln \left(\frac{5}{3}\right)$
$\ln s \tan \mathrm{ds} f \mathmr{and} N a t u r a l L o g a r i t h m {\log}_{e} \left(x\right)$