What is the integral from-3 to 3 of (16)/(x^2-2x-63) ?

The integral from-3 to 3 of (16)/(x^2-2x-63) is -ln(5/2)-ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-3 to 3 of (16)/(x^2-2x-63)

\int_{- 3}^{3} \frac{16}{x ^{2} - 2 x - 63} d x

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

Decimal

- 1.60943 \dots

Solution steps

\int_{- 3}^{3} \frac{16}{x ^{2} - 2 x - 63} d x

Take the constant out:

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

= 16 \cdot \int_{- 3}^{3} \frac{1}{x ^{2} - 2 x - 63} d x

Take the partial fraction of

\frac{1}{x ^{2} - 2 x - 63} : - \frac{1}{16 ( x + 7 )} + \frac{1}{16 ( x - 9 )}

= 16 \cdot \int_{- 3}^{3} - \frac{1}{16 ( x + 7 )} + \frac{1}{16 ( x - 9 )} 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

= 16 (- \int_{- 3}^{3} \frac{1}{16 ( x + 7 )} d x + \int_{- 3}^{3} \frac{1}{16 ( x - 9 )} d x)

\int_{- 3}^{3} \frac{1}{16 ( x + 7 )} d x = \frac{1}{16} ln (\frac{5}{2})

\int_{- 3}^{3} \frac{1}{16 ( x - 9 )} d x = - \frac{1}{16} ln (2)

= 16 (- \frac{1}{16} ln (\frac{5}{2}) - \frac{1}{16} ln (2))

Simplify

16 (- \frac{1}{16} ln (\frac{5}{2}) - \frac{1}{16} ln (2)) : - ln (\frac{5}{2}) - ln (2)

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

\frac{\partial}{\partial x} (- 5 cos (9 x))

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0, y (0) = 6

y = 2 x + 4

7 x - 8 y x^{2} + 1 (\frac{d y}{d x}) = 0

\int 9 x + \frac{1}{9 x} d x

What is the integral from-3 to 3 of (16)/(x^2-2x-63) ?
The integral from-3 to 3 of (16)/(x^2-2x-63) is -ln(5/2)-ln(2)