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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

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

Add a constant to the solution

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

\int_{0}^{\frac{π}{4}} cos^{2} (\frac{x}{3}) d x

\int_{1}^{2} (x^{2} + 2 x) d x

f (x) = x^{2} + x, at x = - 1, 2

\int_{- 2}^{1} (x^{2} - 6 x + 12) d x

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

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