What is the integral from-5 to-3 of x/(sqrt(x^2-9)) ?

The integral from-5 to-3 of x/(sqrt(x^2-9)) is -4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-5 to-3 of x/(sqrt(x^2-9))

\int_{- 5}^{- 3} \frac{x}{x ^{2} - 9} d x

- 4

Solution steps

\int_{- 5}^{- 3} \frac{x}{x ^{2} - 9} d x

Apply u-substitution

: \int_{16}^{0} \frac{1}{2 u} d u

= \int_{16}^{0} \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{16} \frac{1}{2 u} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int_{0}^{16} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{16}

= - \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{16}

Apply radical rule:

a^{\frac{1}{n}} = n a

= - \frac{1}{2} [2 u]_{0}^{16}

Compute the boundaries:

8

= - \frac{1}{2} \cdot 8

Simplify

= - 4

\int_{y}^{\infty} e^{- x} d x

\int_{0}^{x} y d y

\int_{y}^{1} y ln (x) d x

\int_{0}^{2} \frac{1}{x ^{2} + 1} d x

\int_{0}^{1} (3 + x^{2})^{2} x d x

What is the integral from-5 to-3 of x/(sqrt(x^2-9)) ?
The integral from-5 to-3 of x/(sqrt(x^2-9)) is -4