What is the integral from-3 to 3 of (12)/(x^2-2x-35) ?

The integral from-3 to 3 of (12)/(x^2-2x-35) is -2ln(2)+ln(2/5)

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integral from-3 to 3 of (12)/(x^2-2x-35)

\int_{- 3}^{3} \frac{12}{x ^{2} - 2 x - 35} d x

- 2 ln (2) + ln (\frac{2}{5})

Decimal Notation

- 2.30258 \dots

Solution steps

\int_{- 3}^{3} \frac{12}{x ^{2} - 2 x - 35} d x

Take the constant out:

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

= 12 \cdot \int_{- 3}^{3} \frac{1}{x ^{2} - 2 x - 35} d x

Take the partial fraction of

\frac{1}{x ^{2} - 2 x - 35} : - \frac{1}{12 ( x + 5 )} + \frac{1}{12 ( x - 7 )}

= 12 \cdot \int_{- 3}^{3} - \frac{1}{12 ( x + 5 )} + \frac{1}{12 ( x - 7 )} 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_{- 3}^{3} \frac{1}{12 ( x + 5 )} d x + \int_{- 3}^{3} \frac{1}{12 ( x - 7 )} d x)

\int_{- 3}^{3} \frac{1}{12 ( x + 5 )} d x = \frac{1}{6} ln (2)

\int_{- 3}^{3} \frac{1}{12 ( x - 7 )} d x = \frac{1}{12} ln (\frac{2}{5})

= 12 (- \frac{1}{6} ln (2) + \frac{1}{12} ln (\frac{2}{5}))

Simplify

12 (- \frac{1}{6} ln (2) + \frac{1}{12} ln (\frac{2}{5})) : - 2 ln (2) + ln (\frac{2}{5})

= - 2 ln (2) + ln (\frac{2}{5})

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\int \frac{x ^{3}}{( 4 - x ^{4} ) ^{6}} d x

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What is the integral from-3 to 3 of (12)/(x^2-2x-35) ?
The integral from-3 to 3 of (12)/(x^2-2x-35) is -2ln(2)+ln(2/5)