What is the integral of (-x^2-9x-1)/(x+8) ?

The integral of (-x^2-9x-1)/(x+8) is 7x+7ln|x+8|-1/2 (x+8)^2+56+C

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integral of (-x^2-9x-1)/(x+8)

\int \frac{- x ^{2} - 9 x - 1}{x + 8} d x

7 x + 7 ln ∣ x + 8 ∣ - \frac{1}{2} (x + 8)^{2} + 56 + C

Solution steps

\int \frac{- x ^{2} - 9 x - 1}{x + 8} d x

Expand

\frac{- x ^{2} - 9 x - 1}{x + 8} : - \frac{x ^{2}}{x + 8} - \frac{9 x}{x + 8} - \frac{1}{x + 8}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{x ^{2}}{x + 8} d x - \int \frac{9 x}{x + 8} d x - \int \frac{1}{x + 8} d x

\int \frac{x ^{2}}{x + 8} d x = \frac{( x + 8 ) ^{2}}{2} - 16 (x + 8) + 64 ln ∣ x + 8 ∣

\int \frac{9 x}{x + 8} d x = 9 (x + 8 - 8 ln ∣ x + 8 ∣)

\int \frac{1}{x + 8} d x = ln ∣ x + 8 ∣

= - (\frac{( x + 8 ) ^{2}}{2} - 16 (x + 8) + 64 ln ∣ x + 8 ∣) - 9 (x + 8 - 8 ln ∣ x + 8 ∣) - ln ∣ x + 8 ∣

Simplify

- (\frac{( x + 8 ) ^{2}}{2} - 16 (x + 8) + 64 ln ∣ x + 8 ∣) - 9 (x + 8 - 8 ln ∣ x + 8 ∣) - ln ∣ x + 8 ∣ : 7 x + 7 ln ∣ x + 8 ∣ - \frac{1}{2} (x + 8)^{2} + 56

= 7 x + 7 ln ∣ x + 8 ∣ - \frac{1}{2} (x + 8)^{2} + 56

Add a constant to the solution

= 7 x + 7 ln ∣ x + 8 ∣ - \frac{1}{2} (x + 8)^{2} + 56 + C

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What is the integral of (-x^2-9x-1)/(x+8) ?
The integral of (-x^2-9x-1)/(x+8) is 7x+7ln|x+8|-1/2 (x+8)^2+56+C