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

The integral from 1 to 2 of 1/(x^2-4x-5) is -1/6 ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{1}{6} ln (2)

Decimal Notation

- 0.11552 \dots

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 5} : - \frac{1}{6 ( x + 1 )} + \frac{1}{6 ( x - 5 )}

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

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

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

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

= - \frac{1}{6} (ln (3) - ln (2)) + \frac{1}{6} (ln (3) - 2 ln (2))

Simplify

- \frac{1}{6} (ln (3) - ln (2)) + \frac{1}{6} (ln (3) - 2 ln (2)) : - \frac{1}{6} ln (2)

= - \frac{1}{6} ln (2)

\int_{0}^{1} \frac{cos ( x )}{1 + x} d x

\int_{0}^{1} (8 - 12 x)^{2} d x

\int_{20}^{40} 3 (3 x + 4)^{2} d x

\int_{3}^{9} 20 e^{- 0.1 t^{2}} d t

\int_{- \infty}^{\frac{π}{2}} tan (θ) d θ

What is the integral from 1 to 2 of 1/(x^2-4x-5) ?
The integral from 1 to 2 of 1/(x^2-4x-5) is -1/6 ln(2)