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

The integral from 1 to 2 of x/(sqrt(x^2-1)) is square root of 3

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

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

3

Decimal Notation

1.73205 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{3} \frac{1}{2 u} d u

= \int_{0}^{3} \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}^{3} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{3}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{3}

Apply radical rule:

= \frac{1}{2} [2 u]_{0}^{3}

Compute the boundaries:

23

= \frac{1}{2} \cdot 23

Simplify

= 3

\frac{∣ x ∣}{5 - x ^{2}}

x^{2} y d x - (x^{3} + y^{3}) d y = 0

f (x) = (2 x^{2} + 3) (5 x - 4)

f (x) = \frac{e ^{x}}{x}

x \to - 3 lim (\frac{x ^{3} + 8}{x + 2})

What is the integral from 1 to 2 of x/(sqrt(x^2-1)) ?
The integral from 1 to 2 of x/(sqrt(x^2-1)) is square root of 3