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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

\int u d u = \frac{u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

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

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

Substitute back

u = x - 1

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

Add a constant to the solution

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

n = 2 \sum \infty \frac{6}{5 ^{n}}

\int cos^{4} (x s i) n^{3} x d x

\int_{1}^{4} \frac{u - 2}{u} d u

\frac{\partial}{\partial y} ((x^{2} y - y^{3})^{5})

\frac{d y}{d x} = \frac{cos ( 9 x )}{e ^{9 y}}

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