What is the integral of 1/((1-x)(x^3-3x^2+3x-1)) ?

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

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integral of 1/((1-x)(x^3-3x^2+3x-1))

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{( 1 - x ) ( x ^{3} - 3 x ^{2} + 3 x - 1 )} : - \frac{1}{( x - 1 ) ^{4}}

= \int - \frac{1}{( x - 1 ) ^{4}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{( x - 1 ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{u ^{4}} d u

= - \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= - (- \frac{1}{3 u ^{3}})

Substitute back

u = x - 1

= - (- \frac{1}{3 ( x - 1 ) ^{3}})

Simplify

= \frac{1}{3 ( x - 1 ) ^{3}}

Add a constant to the solution

= \frac{1}{3 ( x - 1 ) ^{3}} + C

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