How do you find the indefinite integral of $\int \frac{x^{2} - 6 x - 20}{x + 5}$ $int (x^2-6x-20)/(x+5)$ ?

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Answer 1 · 2017-04-13T14:21:20

The answer is $= \frac{(x + 5) (x - 27)}{2} + 35 ln (| x + 5 |) + C$ $=((x+5)(x-27))/2+35ln(|x+5|)+C$

We perform this integral by substitution.

Let $u = x + 5$ $u=x+5$ , $\Rightarrow$ $=>$ , $d x = d u$ $dx=du$

$x = u - 5$ $x=u-5$

and

$x^{2} - 6 x - 20 = {(u - 5)}^{2} - 6 (u - 5) - 20$ $x^2-6x-20=(u-5)^2-6(u-5)-20$

$= u^{2} - 10 u + 25 - 6 u + 30 - 20$ $=u^2-10u+25-6u+30-20$

$= u^{2} - 16 u + 35$ $=u^2-16u+35$

Therefore,

$\int \frac{(x^{2} - 6 x - 20) d x}{x + 5}$ $int((x^2-6x-20)dx)/(x+5)$

$= \int \frac{(u^{2} - 16 u + 35) d u}{u}$ $=int((u^2-16u+35)du)/u$

$= \int (u - 16 + \frac{35}{u})$ $=int(u-16+35/u)$

$= \frac{u^{2}}{2} - 16 u + 35 ln u$ $=u^2/2-16u+35lnu$

$= \frac{{(x + 5)}^{2}}{2} - 16 (x + 5) + 35 ln (| x + 5 |) + C$ $=(x+5)^2/2-16(x+5)+35ln(|x+5|)+C$

$= (x + 5) \frac{x + 5 - 32}{2} + 35 ln (| x + 5 |) + C$ $=(x+5)(x+5-32)/2+35ln(|x+5|)+C$

$= \frac{(x + 5) (x - 27)}{2} + 35 ln (| x + 5 |) + C$ $=((x+5)(x-27))/2+35ln(|x+5|)+C$

How do you find the indefinite integral of int (x^2-6x-20)/(x+5)∫x2−6x−20x+5int (x^2-6x-20)/(x+5)?