What is the integral from 0 to 1/6 of (x^3)/(x^2-1) ?

The integral from 0 to 1/6 of (x^3)/(x^2-1) is 1/2 (1/36+ln(35/36))

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integral from 0 to 1/6 of (x^3)/(x^2-1)

\int_{0}^{\frac{1}{6}} \frac{x ^{3}}{x ^{2} - 1} d x

\frac{1}{2} (\frac{1}{36} + ln (\frac{35}{36}))

Decimal Notation

- 0.00019 \dots

Solution steps

\int_{0}^{\frac{1}{6}} \frac{x ^{3}}{x ^{2} - 1} d x

Apply u-substitution

: \int_{- 1}^{- \frac{35}{36}} \frac{u + 1}{2 u} d u

= \int_{- 1}^{- \frac{35}{36}} \frac{u + 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}^{- \frac{35}{36}} \frac{u + 1}{u} d u

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

= \frac{1}{2} \cdot \int_{- 1}^{- \frac{35}{36}} 1 + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{- 1}^{- \frac{35}{36}} 1 d u + \int_{- 1}^{- \frac{35}{36}} \frac{1}{u} d u)

\int_{- 1}^{- \frac{35}{36}} 1 d u = \frac{1}{36}

\int_{- 1}^{- \frac{35}{36}} \frac{1}{u} d u = ln (\frac{35}{36})

= \frac{1}{2} (\frac{1}{36} + ln (\frac{35}{36}))

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What is the integral from 0 to 1/6 of (x^3)/(x^2-1) ?
The integral from 0 to 1/6 of (x^3)/(x^2-1) is 1/2 (1/36+ln(35/36))