What is the integral from-1 to 5 of (8x)/(x^2-4x-12) ?

The integral from-1 to 5 of (8x)/(x^2-4x-12) is -4ln(7)

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integral from-1 to 5 of (8x)/(x^2-4x-12)

\int_{- 1}^{5} \frac{8 x}{x ^{2} - 4 x - 12} d x

- 4 ln (7)

Decimal

- 7.78364 \dots

Solution steps

\int_{- 1}^{5} \frac{8 x}{x ^{2} - 4 x - 12} d x

Take the constant out:

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

= 8 \cdot \int_{- 1}^{5} \frac{x}{x ^{2} - 4 x - 12} d x

Take the partial fraction of

\frac{x}{x ^{2} - 4 x - 12} : \frac{1}{4 ( x + 2 )} + \frac{3}{4 ( x - 6 )}

= 8 \cdot \int_{- 1}^{5} \frac{1}{4 ( x + 2 )} + \frac{3}{4 ( 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

= 8 (\int_{- 1}^{5} \frac{1}{4 ( x + 2 )} d x + \int_{- 1}^{5} \frac{3}{4 ( x - 6 )} d x)

\int_{- 1}^{5} \frac{1}{4 ( x + 2 )} d x = \frac{1}{4} ln (7)

\int_{- 1}^{5} \frac{3}{4 ( x - 6 )} d x = - \frac{3}{4} ln (7)

= 8 (\frac{1}{4} ln (7) - \frac{3}{4} ln (7))

Simplify

8 (\frac{1}{4} ln (7) - \frac{3}{4} ln (7)) : - 4 ln (7)

= - 4 ln (7)

What is the integral from-1 to 5 of (8x)/(x^2-4x-12) ?
The integral from-1 to 5 of (8x)/(x^2-4x-12) is -4ln(7)