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

The integral from 2 to 5 of x/(sqrt(x^2-4)) is square root of 21

Popular >

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

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

21

Decimal

4.58257 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{21} \frac{1}{2 u} d u

= \int_{0}^{21} \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}^{21} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{21}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{21}

Apply radical rule:

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

= \frac{1}{2} [2 u]_{0}^{21}

Compute the boundaries:

221

= \frac{1}{2} \cdot 221

Simplify

= 21

f (x) = \frac{- 1}{2 x} + 2 x

\frac{d}{d x} (cos (x^{- 4}))

x \to 2 lim (\frac{5}{x - 2})

(3, - 3), (5, - 6)

y^{^{''}} - 2 y^{^{'}} + y = x e^{x} ln (x)

What is the integral from 2 to 5 of x/(sqrt(x^2-4)) ?
The integral from 2 to 5 of x/(sqrt(x^2-4)) is square root of 21