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

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

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

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

\frac{1}{2} ln (3)

Decimal Notation

0.54930 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= \frac{1}{2} ln (3)

\int 3 x x^{2} - 1 d x

\frac{d P}{d t} = P (0.1 - 1 0^{- 7} P)

\int (e^{2} - 2^{e}) e^{x} d x

z \to 2 i lim (z^{2})

\int x^{3} x^{2} + 2 d x

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