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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \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 )}

= 3 \cdot \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

= 3 (\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 ∣

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

Simplify

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

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

Add a constant to the solution

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

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