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

The integral of (x^4+x^3+x^2+1)/(x^2+x-2) is (x^3)/3+3x+4/3 ln|x-1|-13/3 ln|x+2|+C

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integral of (x^4+x^3+x^2+1)/(x^2+x-2)

\int \frac{x ^{4} + x ^{3} + x ^{2} + 1}{x ^{2} + x - 2} d x

\frac{x ^{3}}{3} + 3 x + \frac{4}{3} ln ∣ x - 1 ∣ - \frac{13}{3} ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{x ^{4} + x ^{3} + x ^{2} + 1}{x ^{2} + x - 2} d x

Long division

\frac{x ^{4} + x ^{3} + x ^{2} + 1}{x ^{2} + x - 2} : x^{2} + 3 + \frac{- 3 x + 7}{x ^{2} + x - 2}

= \int x^{2} + 3 + \frac{- 3 x + 7}{x ^{2} + x - 2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x + \int 3 d x + \int \frac{- 3 x + 7}{x ^{2} + x - 2} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 3 d x = 3 x

\int \frac{- 3 x + 7}{x ^{2} + x - 2} d x = \frac{4}{3} ln ∣ x - 1 ∣ - \frac{13}{3} ln ∣ x + 2 ∣

= \frac{x ^{3}}{3} + 3 x + \frac{4}{3} ln ∣ x - 1 ∣ - \frac{13}{3} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + 3 x + \frac{4}{3} ln ∣ x - 1 ∣ - \frac{13}{3} ln ∣ x + 2 ∣ + C

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

\int \frac{8}{36 x ^{2} + 4} d x

\frac{d}{d x} (e^{- x} cos (\frac{x}{2}))

\frac{\partial}{\partial v} (3 u sin (v))

\frac{d y}{d x} + y^{3} x + 6 y = 0

What is the integral of (x^4+x^3+x^2+1)/(x^2+x-2) ?
The integral of (x^4+x^3+x^2+1)/(x^2+x-2) is (x^3)/3+3x+4/3 ln|x-1|-13/3 ln|x+2|+C