What is the integral from-3 to 3 of (16)/(x^2-6x-55) ?

The integral from-3 to 3 of (16)/(x^2-6x-55) is -2ln(2)+ln(4/7)

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^2-6x-55)

\int_{- 3}^{3} \frac{16}{x ^{2} - 6 x - 55} d x

- 2 ln (2) + ln (\frac{4}{7})

Decimal Notation

- 1.94591 \dots

Solution steps

\int_{- 3}^{3} \frac{16}{x ^{2} - 6 x - 55} 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} - 6 x - 55} d x

Take the partial fraction of

\frac{1}{x ^{2} - 6 x - 55} : - \frac{1}{16 ( x + 5 )} + \frac{1}{16 ( x - 11 )}

= 16 \cdot \int_{- 3}^{3} - \frac{1}{16 ( x + 5 )} + \frac{1}{16 ( x - 11 )} 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 + 5 )} d x + \int_{- 3}^{3} \frac{1}{16 ( x - 11 )} d x)

\int_{- 3}^{3} \frac{1}{16 ( x + 5 )} d x = \frac{1}{8} ln (2)

\int_{- 3}^{3} \frac{1}{16 ( x - 11 )} d x = \frac{1}{16} ln (\frac{4}{7})

= 16 (- \frac{1}{8} ln (2) + \frac{1}{16} ln (\frac{4}{7}))

Simplify

16 (- \frac{1}{8} ln (2) + \frac{1}{16} ln (\frac{4}{7})) : - 2 ln (2) + ln (\frac{4}{7})

= - 2 ln (2) + ln (\frac{4}{7})

\frac{d}{d x} ((1 - e^{x}) (x + e^{x}))

\frac{\partial}{\partial z} (x^{2} y z^{2} + sin (yz))

\int 1 + \frac{x ^{2}}{2 - x ^{2}} d x

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

x \to 0 lim (15)

What is the integral from-3 to 3 of (16)/(x^2-6x-55) ?
The integral from-3 to 3 of (16)/(x^2-6x-55) is -2ln(2)+ln(4/7)