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

The integral of (-x+8)/(-x^2-x+8) is

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

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

\frac{1}{2} ln - 4 x^{2} - 4 x + 32 + \frac{17}{2 33} ln \frac{1}{33} (2 x + 1) + 1 - \frac{17}{2 33} ln \frac{1}{33} (2 x + 1) - 1 + C

Solution steps

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

Expand

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

\int \frac{x}{- x ^{2} - x + 8} d x = 2 (- \frac{1}{4} ln - 4 x^{2} - 4 x + 32 - \frac{1}{4 33} (ln \frac{2 x + 1}{33} + 1 - ln \frac{2 x + 1}{33} - 1))

\int \frac{8}{- x ^{2} - x + 8} d x = \frac{8}{33} (ln \frac{2 x + 1}{33} + 1 - ln \frac{2 x + 1}{33} - 1)

= - 2 (- \frac{1}{4} ln - 4 x^{2} - 4 x + 32 - \frac{1}{4 33} (ln \frac{2 x + 1}{33} + 1 - ln \frac{2 x + 1}{33} - 1)) + \frac{8}{33} (ln \frac{2 x + 1}{33} + 1 - ln \frac{2 x + 1}{33} - 1)

Simplify

- 2 (- \frac{1}{4} ln - 4 x^{2} - 4 x + 32 - \frac{1}{4 33} (ln \frac{2 x + 1}{33} + 1 - ln \frac{2 x + 1}{33} - 1)) + \frac{8}{33} (ln \frac{2 x + 1}{33} + 1 - ln \frac{2 x + 1}{33} - 1) : \frac{1}{2} ln - 4 x^{2} - 4 x + 32 + \frac{17}{2 33} ln \frac{1}{33} (2 x + 1) + 1 - \frac{17}{2 33} ln \frac{1}{33} (2 x + 1) - 1

= \frac{1}{2} ln - 4 x^{2} - 4 x + 32 + \frac{17}{2 33} ln \frac{1}{33} (2 x + 1) + 1 - \frac{17}{2 33} ln \frac{1}{33} (2 x + 1) - 1

Add a constant to the solution

= \frac{1}{2} ln - 4 x^{2} - 4 x + 32 + \frac{17}{2 33} ln \frac{1}{33} (2 x + 1) + 1 - \frac{17}{2 33} ln \frac{1}{33} (2 x + 1) - 1 + C

x \to 1 lim (\frac{1 + x}{1 - x})

\frac{1}{1 + 1 x}

y^{^{'}} + t y = t

x^{2} - 3

y^{^{'}} + \frac{1}{x} y = \frac{1}{x ^{2}}

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