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

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

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integral from 2 to 3 of x/(sqrt(x^2-4))

\int_{2}^{3} \frac{x}{x ^{2} - 4} d x

5

Decimal Notation

2.23606 \dots

Solution steps

\int_{2}^{3} \frac{x}{x ^{2} - 4} d x

Apply u-substitution

: \int_{0}^{5} \frac{1}{2 u} d u

= \int_{0}^{5} \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}^{5} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{5}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{5}

Apply radical rule:

= \frac{1}{2} [2 u]_{0}^{5}

Compute the boundaries:

25

= \frac{1}{2} \cdot 25

Simplify

= 5

\int_{0}^{6} 5 x^{2} + 10 x + 6 d x

\int_{0}^{3} - 4 x^{2} d x

\int_{2}^{6} \frac{1}{x} d x

\int_{0}^{1} 2 e^{6 x - 1} d x

\int_{10}^{x} \frac{- 1}{75} (x - 20) d x

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