What is the integral from-3 to 3 of (10)/(x^2-2x-24) ?

The integral from-3 to 3 of (10)/(x^2-2x-24) is -ln(7)-ln(3)

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integral from-3 to 3 of (10)/(x^2-2x-24)

\int_{- 3}^{3} \frac{10}{x ^{2} - 2 x - 24} d x

- ln (7) - ln (3)

Decimal

- 3.04452 \dots

Solution steps

\int_{- 3}^{3} \frac{10}{x ^{2} - 2 x - 24} d x

Take the constant out:

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

= 10 \cdot \int_{- 3}^{3} \frac{1}{x ^{2} - 2 x - 24} d x

Take the partial fraction of

\frac{1}{x ^{2} - 2 x - 24} : - \frac{1}{10 ( x + 4 )} + \frac{1}{10 ( x - 6 )}

= 10 \cdot \int_{- 3}^{3} - \frac{1}{10 ( x + 4 )} + \frac{1}{10 ( 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

= 10 (- \int_{- 3}^{3} \frac{1}{10 ( x + 4 )} d x + \int_{- 3}^{3} \frac{1}{10 ( x - 6 )} d x)

\int_{- 3}^{3} \frac{1}{10 ( x + 4 )} d x = \frac{1}{10} ln (7)

\int_{- 3}^{3} \frac{1}{10 ( x - 6 )} d x = - \frac{1}{10} ln (3)

= 10 (- \frac{1}{10} ln (7) - \frac{1}{10} ln (3))

Simplify

10 (- \frac{1}{10} ln (7) - \frac{1}{10} ln (3)) : - ln (7) - ln (3)

= - ln (7) - ln (3)

\frac{d}{d x} (2 x^{3} - 3 x^{2} + 2)

\int \frac{1}{x ( x - 0.5 )} d x

\frac{d}{d x} (x^{2} - 9 x)

\int_{0}^{\frac{π}{2}} 3 cos^{2} (x) d x

\frac{d}{d x} (x ln (x) + (3 x^{2} - 2)^{3})

What is the integral from-3 to 3 of (10)/(x^2-2x-24) ?
The integral from-3 to 3 of (10)/(x^2-2x-24) is -ln(7)-ln(3)