What is the integral from 3 to 4 of x/(x^2-1) ?

The integral from 3 to 4 of x/(x^2-1) is 1/2 (ln(15)-3ln(2))

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integral from 3 to 4 of x/(x^2-1)

\int_{3}^{4} \frac{x}{x ^{2} - 1} d x

\frac{1}{2} (ln (15) - 3 ln (2))

Decimal Notation

0.31430 \dots

Solution steps

\int_{3}^{4} \frac{x}{x ^{2} - 1} d x

Apply u-substitution

: \int_{8}^{15} \frac{1}{2 u} d u

= \int_{8}^{15} \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_{8}^{15} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{8}^{15}

Compute the boundaries:

ln (15) - 3 ln (2)

= \frac{1}{2} (ln (15) - 3 ln (2))

\int \frac{1}{( x - 3 )} d x

\frac{y}{x}

\frac{\partial}{\partial x} (2 x e^{x y})

9 z^{2} e^{z}

\int x^{5} ln (8 x) d x

What is the integral from 3 to 4 of x/(x^2-1) ?
The integral from 3 to 4 of x/(x^2-1) is 1/2 (ln(15)-3ln(2))