What is the integral from-2 to 2 of (12)/(x^2-4x-32) ?

The integral from-2 to 2 of (12)/(x^2-4x-32) is -ln(3)+ln(3/5)

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integral from-2 to 2 of (12)/(x^2-4x-32)

\int_{- 2}^{2} \frac{12}{x ^{2} - 4 x - 32} d x

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

Decimal Notation

- 1.60943 \dots

Solution steps

\int_{- 2}^{2} \frac{12}{x ^{2} - 4 x - 32} d x

Take the constant out:

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

= 12 \cdot \int_{- 2}^{2} \frac{1}{x ^{2} - 4 x - 32} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 32} : - \frac{1}{12 ( x + 4 )} + \frac{1}{12 ( x - 8 )}

= 12 \cdot \int_{- 2}^{2} - \frac{1}{12 ( x + 4 )} + \frac{1}{12 ( x - 8 )} 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

= 12 (- \int_{- 2}^{2} \frac{1}{12 ( x + 4 )} d x + \int_{- 2}^{2} \frac{1}{12 ( x - 8 )} d x)

\int_{- 2}^{2} \frac{1}{12 ( x + 4 )} d x = \frac{1}{12} ln (3)

\int_{- 2}^{2} \frac{1}{12 ( x - 8 )} d x = \frac{1}{12} ln (\frac{3}{5})

= 12 (- \frac{1}{12} ln (3) + \frac{1}{12} ln (\frac{3}{5}))

Simplify

12 (- \frac{1}{12} ln (3) + \frac{1}{12} ln (\frac{3}{5})) : - ln (3) + ln (\frac{3}{5})

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

\frac{d}{d x} (sin^{2} (x^{3}))

4 x lo g_{10} (x^{9})

\int \frac{5 x ^{7} - 9 x ^{5}}{x ^{3}} d x

f (x) = \frac{2 x}{2 + x}

\frac{\partial}{\partial x} (4 x cos (y))

What is the integral from-2 to 2 of (12)/(x^2-4x-32) ?
The integral from-2 to 2 of (12)/(x^2-4x-32) is -ln(3)+ln(3/5)