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

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

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

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

ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ + x - 1 + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

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

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

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

\int 1 d u = u

= ln ∣ u + 1 ∣ - 4 ln ∣ u + 2 ∣ + u

Substitute back

u = x - 1

= ln ∣ x - 1 + 1 ∣ - 4 ln ∣ x - 1 + 2 ∣ + x - 1

Simplify

= ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ + x - 1

Add a constant to the solution

= ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ + x - 1 + C

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