What is the integral from-2 to 2 of (16)/(x^2-4x-60) ?

The integral from-2 to 2 of (16)/(x^2-4x-60) is -ln(2)+ln(2/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to 2 of (16)/(x^2-4x-60)

\int_{- 2}^{2} \frac{16}{x ^{2} - 4 x - 60} d x

- ln (2) + ln (\frac{2}{3})

Decimal Notation

- 1.09861 \dots

Solution steps

\int_{- 2}^{2} \frac{16}{x ^{2} - 4 x - 60} d x

Take the constant out:

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

= 16 \cdot \int_{- 2}^{2} \frac{1}{x ^{2} - 4 x - 60} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 60} : - \frac{1}{16 ( x + 6 )} + \frac{1}{16 ( x - 10 )}

= 16 \cdot \int_{- 2}^{2} - \frac{1}{16 ( x + 6 )} + \frac{1}{16 ( x - 10 )} 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_{- 2}^{2} \frac{1}{16 ( x + 6 )} d x + \int_{- 2}^{2} \frac{1}{16 ( x - 10 )} d x)

\int_{- 2}^{2} \frac{1}{16 ( x + 6 )} d x = \frac{1}{16} ln (2)

\int_{- 2}^{2} \frac{1}{16 ( x - 10 )} d x = \frac{1}{16} ln (\frac{2}{3})

= 16 (- \frac{1}{16} ln (2) + \frac{1}{16} ln (\frac{2}{3}))

Simplify

16 (- \frac{1}{16} ln (2) + \frac{1}{16} ln (\frac{2}{3})) : - ln (2) + ln (\frac{2}{3})

= - ln (2) + ln (\frac{2}{3})

y^{^{'''}} + y^{^{'}} = x

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

\frac{7}{s ^{2} + 42}

\frac{d}{d x} (1 - ln^{2} (x))

\frac{\partial}{\partial x} (x^{2} - 7 x y + y)

What is the integral from-2 to 2 of (16)/(x^2-4x-60) ?
The integral from-2 to 2 of (16)/(x^2-4x-60) is -ln(2)+ln(2/3)