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

The integral of (x^2)/(x^2-2x+1) is (x^3)/(x^2-2x+1)+2ln|x-1|-4/(x-1)-1/((x-1)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{x ^{3}}{x ^{2} - 2 x + 1} + 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{2 x ^{2}}{( x - 1 ) ^{3}} d x = - 2 ln ∣ x - 1 ∣ + \frac{4}{x - 1} + \frac{1}{( x - 1 ) ^{2}}

= \frac{x ^{3}}{x ^{2} - 2 x + 1} - (- 2 ln ∣ x - 1 ∣ + \frac{4}{x - 1} + \frac{1}{( x - 1 ) ^{2}})

Simplify

= \frac{x ^{3}}{x ^{2} - 2 x + 1} + 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

Add a constant to the solution

= \frac{x ^{3}}{x ^{2} - 2 x + 1} + 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

\frac{1}{8 x ^{2}}

n = 1 \sum \infty 3^{n}

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

n \to \infty lim (3)

\frac{\partial}{\partial x} (cos (y e^{x y}) sin (x))

What is the integral of (x^2)/(x^2-2x+1) ?
The integral of (x^2)/(x^2-2x+1) is (x^3)/(x^2-2x+1)+2ln|x-1|-4/(x-1)-1/((x-1)^2)+C