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

The integral of ((x-1)^3)/(x^3+(x-1)^3) is 1/2 x-1/12 ln|2x-1|-1/3 ln|x^2-x+1|+C

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integral of ((x-1)^3)/(x^3+(x-1)^3)

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

\frac{1}{2} x - \frac{1}{12} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln x^{2} - x + 1 + C

Solution steps

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

Take the partial fraction of

\frac{( x - 1 ) ^{3}}{x ^{3} + ( x - 1 ) ^{3}} : \frac{1}{2} - \frac{1}{6 ( 2 x - 1 )} + \frac{- 2 x + 1}{3 ( x ^{2} - x + 1 )}

= \int \frac{1}{2} - \frac{1}{6 ( 2 x - 1 )} + \frac{- 2 x + 1}{3 ( x ^{2} - x + 1 )} d x

Apply the Sum Rule:

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

= \int \frac{1}{2} d x - \int \frac{1}{6 ( 2 x - 1 )} d x + \int \frac{- 2 x + 1}{3 ( x ^{2} - x + 1 )} d x

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{1}{6 ( 2 x - 1 )} d x = \frac{1}{12} ln ∣ 2 x - 1 ∣

\int \frac{- 2 x + 1}{3 ( x ^{2} - x + 1 )} d x = - \frac{1}{3} ln x^{2} - x + 1

= \frac{1}{2} x - \frac{1}{12} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln x^{2} - x + 1

Add a constant to the solution

= \frac{1}{2} x - \frac{1}{12} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln x^{2} - x + 1 + C

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What is the integral of ((x-1)^3)/(x^3+(x-1)^3) ?
The integral of ((x-1)^3)/(x^3+(x-1)^3) is 1/2 x-1/12 ln|2x-1|-1/3 ln|x^2-x+1|+C