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

The integral from 0 to 1 of ((x-1)/(x+1))^3 is (-96ln(2)+64)/(16)

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

\int_{0}^{1} (\frac{x - 1}{x + 1})^{3} d x

\frac{- 96 ln ( 2 ) + 64}{16}

Decimal Notation

- 0.15888 \dots

Solution steps

\int_{0}^{1} (\frac{x - 1}{x + 1})^{3} d x

Expand

(\frac{x - 1}{x + 1})^{3} : \frac{x ^{3} - 3 x ^{2} + 3 x - 1}{x ^{3} + 3 x ^{2} + 3 x + 1}

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

Expand

\frac{x ^{3} - 3 x ^{2} + 3 x - 1}{x ^{3} + 3 x ^{2} + 3 x + 1} : \frac{x ^{3}}{x ^{3} + 3 x ^{2} + 3 x + 1} - \frac{3 x ^{2}}{x ^{3} + 3 x ^{2} + 3 x + 1} + \frac{3 x}{x ^{3} + 3 x ^{2} + 3 x + 1} - \frac{1}{x ^{3} + 3 x ^{2} + 3 x + 1}

Apply the Sum Rule:

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

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

\int_{0}^{1} \frac{x ^{3}}{x ^{3} + 3 x ^{2} + 3 x + 1} d x = \frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2}

\int_{0}^{1} \frac{3 x ^{2}}{x ^{3} + 3 x ^{2} + 3 x + 1} d x = 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2})

\int_{0}^{1} \frac{3 x}{x ^{3} + 3 x ^{2} + 3 x + 1} d x = \frac{3}{8}

\int_{0}^{1} \frac{1}{x ^{3} + 3 x ^{2} + 3 x + 1} d x = \frac{3}{8}

= \frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2} - 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}) + \frac{3}{8} - \frac{3}{8}

Simplify

\frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2} - 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}) + \frac{3}{8} - \frac{3}{8} : \frac{- 96 ln ( 2 ) + 64}{16}

= \frac{- 96 ln ( 2 ) + 64}{16}

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What is the integral from 0 to 1 of ((x-1)/(x+1))^3 ?
The integral from 0 to 1 of ((x-1)/(x+1))^3 is (-96ln(2)+64)/(16)