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

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

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

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

- 3 ln (2) + 2

Decimal

- 0.07944 \dots

Solution steps

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

Complete the square

x^{2} - 2 x + 1 : (x - 1)^{2}

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

Apply u-substitution

: \int_{- 2}^{- 1} \frac{( u + 1 ) ^{3}}{u ^{2}} d u

= \int_{- 2}^{- 1} \frac{( u + 1 ) ^{3}}{u ^{2}} d u

Expand

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

= \int_{- 2}^{- 1} u + 3 + \frac{3}{u} + \frac{1}{u ^{2}} 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

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

\int_{- 2}^{- 1} u d u = - \frac{3}{2}

\int_{- 2}^{- 1} 3 d u = 3

\int_{- 2}^{- 1} \frac{3}{u} d u = - 3 ln (2)

\int_{- 2}^{- 1} \frac{1}{u ^{2}} d u = \frac{1}{2}

= - \frac{3}{2} + 3 - 3 ln (2) + \frac{1}{2}

Simplify

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

= - 3 ln (2) + 2

\int \frac{1}{( 6 - y ^{2} ) ^{\frac{3}{2}}} d y

x^{3} y^{^{'''}} - x^{2} y^{^{''}} + 2 x y^{^{'}} - 2 y = x^{3}

\int \frac{1}{x ^{2} + \frac{3}{4}} d x

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

f (x) = - x^{3}, (- 1, 1)

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