What is the integral from-3 to 3 of (18)/(x^2-6x-72) ?

The integral from-3 to 3 of (18)/(x^2-6x-72) is -ln(3)+ln(3/5)

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integral from-3 to 3 of (18)/(x^2-6x-72)

\int_{- 3}^{3} \frac{18}{x ^{2} - 6 x - 72} d x

- ln (3) + ln (\frac{3}{5})

Decimal

- 1.60943 \dots

Solution steps

\int_{- 3}^{3} \frac{18}{x ^{2} - 6 x - 72} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \int_{- 3}^{3} \frac{1}{x ^{2} - 6 x - 72} d x

Take the partial fraction of

\frac{1}{x ^{2} - 6 x - 72} : - \frac{1}{18 ( x + 6 )} + \frac{1}{18 ( x - 12 )}

= 18 \cdot \int_{- 3}^{3} - \frac{1}{18 ( x + 6 )} + \frac{1}{18 ( x - 12 )} 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

= 18 (- \int_{- 3}^{3} \frac{1}{18 ( x + 6 )} d x + \int_{- 3}^{3} \frac{1}{18 ( x - 12 )} d x)

\int_{- 3}^{3} \frac{1}{18 ( x + 6 )} d x = \frac{1}{18} ln (3)

\int_{- 3}^{3} \frac{1}{18 ( x - 12 )} d x = \frac{1}{18} ln (\frac{3}{5})

= 18 (- \frac{1}{18} ln (3) + \frac{1}{18} ln (\frac{3}{5}))

Simplify

18 (- \frac{1}{18} ln (3) + \frac{1}{18} ln (\frac{3}{5})) : - ln (3) + ln (\frac{3}{5})

= - ln (3) + ln (\frac{3}{5})

\frac{\partial}{\partial y} (\frac{2 x}{( 2 x + 3 y ) ^{2}})

f (x) = e^{t}

ln ∣ sin (x) ∣

y = 2 x + 1, (4, 3)

2 t^{- \frac{1}{4}}

What is the integral from-3 to 3 of (18)/(x^2-6x-72) ?
The integral from-3 to 3 of (18)/(x^2-6x-72) is -ln(3)+ln(3/5)