How do you evaluate $\int \frac{1}{x + 7} (x^{2} + 9)$ $int 1/(x+7)(x^2+9)$ for [-1, 3]?

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How do you evaluate $\int \frac{1}{x + 7} (x^{2} + 9)$ $int 1/(x+7)(x^2+9)$ for [-1, 3]?

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Answer 1 · 2015-09-16T14:07:55

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

Explanation:

the given $\frac{x^{2} + 9}{x + 7}$ $(x^2+9)/(x+7)$ can be written as $x + \frac{- 7 x + 9}{x + 7}$ $x+(-7x+9)/(x+7)$
which can be furthur reduced as $x - 7 + \frac{58}{x + 7}$ $x-7+58/(x+7)$
$\int_{- 1}^{3}$ $int_ -1^3$ $x - 7 + \frac{58}{x + 7} d x$ $x-7+58/(x+7)dx$ = $\int_{- 1}^{3}$ $int_-1^3$ $\frac{x^{2} + 9}{x + 7}$ $(x^2+9)/(x+7)$ $d x$ $dx$
$\int_{- 1}^{3}$ $int_ -1^3$ $x d x - 7 d x + \frac{58}{x + 7} d x$ $xdx-7dx+58/(x+7)dx$
$\frac{x^{2}}{2} - 7 x + 58 ln (x + 7)$ $x^2/2-7x+58ln(x+7)$ from $[- 13]$ $[-1 3]$
$\frac{9}{2} - 7 \cdot 3 + 58 ln 10$ $9/2-7*3+58ln10$ - $(\frac{1}{2} - 7 \cdot (- 1) + 58 ln 6)$ $(1/2-7*(-1)+58ln6)$
$4 - 28 + 58 ln (\frac{10}{6})$ $4-28+58ln(10/6)$
$- 24 + 58 ln (\frac{5}{3})$ $-24+58ln(5/3)$
$ln s tan d s f or N a t u r a l log a r i t h m {log}_{e} (x)$ $ln stands for Natural Logarithm log_e(x)$

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How do you evaluate $\int \frac{1}{x + 7} (x^{2} + 9)$ $int 1/(x+7)(x^2+9)$ for [-1, 3]?

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Explanation:

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How do you evaluate ∫1x+7(x2+9)int 1/(x+7)(x^2+9) for [-1, 3]?

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How do you evaluate $\int \frac{1}{x + 7} (x^{2} + 9)$ $int 1/(x+7)(x^2+9)$ for [-1, 3]?