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

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

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

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

- ln (3) - ln (2)

Decimal Notation

- 1.79175 \dots

Solution steps

\int_{- 2}^{2} \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_{- 2}^{2} \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_{- 2}^{2} - \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_{- 2}^{2} \frac{1}{10 ( x + 4 )} d x + \int_{- 2}^{2} \frac{1}{10 ( x - 6 )} d x)

\int_{- 2}^{2} \frac{1}{10 ( x + 4 )} d x = \frac{1}{10} ln (3)

\int_{- 2}^{2} \frac{1}{10 ( x - 6 )} d x = - \frac{1}{10} ln (2)

= 10 (- \frac{1}{10} ln (3) - \frac{1}{10} ln (2))

Simplify

10 (- \frac{1}{10} ln (3) - \frac{1}{10} ln (2)) : - ln (3) - ln (2)

= - ln (3) - ln (2)

\int - e^{x} d x

f (x) = 1 - 2 cos (x)

\int \frac{2 x ^{3}}{3} d x

\int cot (\frac{x}{9}) d x

\frac{x}{( x ^{2} + 3 x ) ^{5}}

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