What is the integral from 6 to 7 of x/(x^2-1) ?

The integral from 6 to 7 of x/(x^2-1) is 1/2 (ln(48)-ln(35))

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integral from 6 to 7 of x/(x^2-1)

\int_{6}^{7} \frac{x}{x ^{2} - 1} d x

\frac{1}{2} (ln (48) - ln (35))

Decimal Notation

0.15792 \dots

Solution steps

\int_{6}^{7} \frac{x}{x ^{2} - 1} d x

Apply u-substitution

: \int_{35}^{48} \frac{1}{2 u} d u

= \int_{35}^{48} \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_{35}^{48} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{35}^{48}

Compute the boundaries:

ln (48) - ln (35)

= \frac{1}{2} (ln (48) - ln (35))

(1 + ln (x) + \frac{y}{x}) d x = (2 - ln (x)) d y

\frac{\partial}{\partial x} (e^{- 3 x - y})

\frac{d}{d x} (arccos (\frac{11}{x}))

\int \frac{8 e ^{3 t}}{1 + e ^{6 t}} d t

\int \frac{x}{x ^{2} + 5} d x

What is the integral from 6 to 7 of x/(x^2-1) ?
The integral from 6 to 7 of x/(x^2-1) is 1/2 (ln(48)-ln(35))