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

The integral of ((x^2-3x+1))/(x^2+x) is x-3/2-5ln|2x+2|+ln(2|x|)+4+C

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

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

x - \frac{3}{2} - 5 ln ∣ 2 x + 2 ∣ + ln (2 ∣ x ∣) + 4 + C

Solution steps

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

Complete the square

x^{2} - 3 x + 1 : (x - \frac{3}{2})^{2} - \frac{5}{4}

= \int \frac{( x - \frac{3}{2} ) ^{2} - \frac{5}{4}}{x ^{2} + x} d x

Apply u-substitution

: \int \frac{4 u ^{2} - 5}{4 u ^{2} + 16 u + 15} d u

= \int \frac{4 u ^{2} - 5}{4 u ^{2} + 16 u + 15} d u

Long division

\frac{4 u ^{2} - 5}{4 u ^{2} + 16 u + 15} : 1 + \frac{- 16 u - 20}{4 u ^{2} + 16 u + 15}

= \int 1 + \frac{- 16 u - 20}{4 u ^{2} + 16 u + 15} 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 1 d u + \int \frac{- 16 u - 20}{4 u ^{2} + 16 u + 15} d u

\int 1 d u = u

\int \frac{- 16 u - 20}{4 u ^{2} + 16 u + 15} d u = - 5 ln ∣ 2 u + 5 ∣ + ln ∣ 2 u + 3 ∣ + 4

= u - 5 ln ∣ 2 u + 5 ∣ + ln ∣ 2 u + 3 ∣ + 4

Substitute back

u = x - \frac{3}{2}

= x - \frac{3}{2} - 5 ln 2 (x - \frac{3}{2}) + 5 + ln 2 (x - \frac{3}{2}) + 3 + 4

Simplify

x - \frac{3}{2} - 5 ln 2 (x - \frac{3}{2}) + 5 + ln 2 (x - \frac{3}{2}) + 3 + 4 : x - \frac{3}{2} - 5 ln ∣ 2 x + 2 ∣ + ln (2 ∣ x ∣) + 4

= x - \frac{3}{2} - 5 ln ∣ 2 x + 2 ∣ + ln (2 ∣ x ∣) + 4

Add a constant to the solution

= x - \frac{3}{2} - 5 ln ∣ 2 x + 2 ∣ + ln (2 ∣ x ∣) + 4 + C

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What is the integral of ((x^2-3x+1))/(x^2+x) ?
The integral of ((x^2-3x+1))/(x^2+x) is x-3/2-5ln|2x+2|+ln(2|x|)+4+C