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

The integral of (x^3)/(x^2+x-2) is (x^2)/2-x+1/3 ln|x-1|+8/3 ln|x+2|+C

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

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

\frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3}}{x ^{2} + x - 2} : x - 1 + \frac{1}{3 ( x - 1 )} + \frac{8}{3 ( x + 2 )}

= \int x - 1 + \frac{1}{3 ( x - 1 )} + \frac{8}{3 ( 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 d x - \int 1 d x + \int \frac{1}{3 ( x - 1 )} d x + \int \frac{8}{3 ( x + 2 )} d x

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

\int 1 d x = x

\int \frac{1}{3 ( x - 1 )} d x = \frac{1}{3} ln ∣ x - 1 ∣

\int \frac{8}{3 ( x + 2 )} d x = \frac{8}{3} ln ∣ x + 2 ∣

= \frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣ + C

\int \frac{x ^{4}}{x ^{5} - 2} d x

\frac{d}{d x} (\frac{e ^{\frac{1}{x}}}{x ^{2}})

x \to \infty lim (1 - \frac{3}{x})

\frac{d}{d y} (- x y)

\frac{d y}{d x} = \frac{x ^{5}}{3 y ^{2}}

What is the integral of (x^3)/(x^2+x-2) ?
The integral of (x^3)/(x^2+x-2) is (x^2)/2-x+1/3 ln|x-1|+8/3 ln|x+2|+C