What is the integral of (x^3+2x^2-2x-5)/(x^2+2x-3) ?

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

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integral of (x^3+2x^2-2x-5)/(x^2+2x-3)

\int \frac{x ^{3} + 2 x ^{2} - 2 x - 5}{x ^{2} + 2 x - 3} d x

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

Solution steps

\int \frac{x ^{3} + 2 x ^{2} - 2 x - 5}{x ^{2} + 2 x - 3} d x

Complete the square

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

= \int \frac{x ^{3} + 2 x ^{2} - 2 x - 5}{( x + 1 ) ^{2} - 4} d x

Apply u-substitution

: \int \frac{- 2 u + ( u - 1 ) ^{3} + 2 ( u - 1 ) ^{2} - 3}{u ^{2} - 4} d u

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

Long division

\frac{- 2 u + ( u - 1 ) ^{3} + 2 ( u - 1 ) ^{2} - 3}{u ^{2} - 4} : u - 1 + \frac{u - 6}{u ^{2} - 4}

= \int u - 1 + \frac{u - 6}{u ^{2} - 4} 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 1 d u + \int \frac{u - 6}{u ^{2} - 4} d u

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

\int 1 d u = u

\int \frac{u - 6}{u ^{2} - 4} d u = 2 ln ∣ u + 2 ∣ - ln ∣ u - 2 ∣

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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