What is the integral from 0 to 5 of (x-3)/(x^2-5x-6) ?

The integral from 0 to 5 of (x-3)/(x^2-5x-6) is 1/7 ln(6)

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integral from 0 to 5 of (x-3)/(x^2-5x-6)

\int_{0}^{5} \frac{x - 3}{x ^{2} - 5 x - 6} d x

\frac{1}{7} ln (6)

Solution steps

\int_{0}^{5} \frac{x - 3}{x ^{2} - 5 x - 6} d x

Take the partial fraction of

\frac{x - 3}{x ^{2} - 5 x - 6} : \frac{4}{7 ( x + 1 )} + \frac{3}{7 ( x - 6 )}

= \int_{0}^{5} \frac{4}{7 ( x + 1 )} + \frac{3}{7 ( x - 6 )} d x

Apply the Sum Rule:

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

= \int_{0}^{5} \frac{4}{7 ( x + 1 )} d x + \int_{0}^{5} \frac{3}{7 ( x - 6 )} d x

\int_{0}^{5} \frac{4}{7 ( x + 1 )} d x = \frac{4}{7} ln (6)

\int_{0}^{5} \frac{3}{7 ( x - 6 )} d x = - \frac{3}{7} ln (6)

= \frac{4}{7} ln (6) - \frac{3}{7} ln (6)

Simplify

\frac{4}{7} ln (6) - \frac{3}{7} ln (6) : \frac{1}{7} ln (6)

= \frac{1}{7} ln (6)

\frac{\partial}{\partial x} (sin (x) ln (x^{4} + \frac{1}{x}))

e^{y} + y cos (x) + (x e^{y} + sin (x)) \frac{d y}{d x} = 0

\int 4 x^{3} - 3 x^{2} + 6 x - 1 d x

e^{x} - e^{- x}

\frac{d}{d x} (e^{5 x y})

What is the integral from 0 to 5 of (x-3)/(x^2-5x-6) ?
The integral from 0 to 5 of (x-3)/(x^2-5x-6) is 1/7 ln(6)