What is the integral of 1/(x^3-x^2-x-15) ?

The integral of 1/(x^3-x^2-x-15) is 1/20 ln|x-3|+1/40 (-ln|x^2+2x+5|-4arctan(1/2 (x+1)))+C

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integral of 1/(x^3-x^2-x-15)

\int \frac{1}{x ^{3} - x ^{2} - x - 15} d x

\frac{1}{20} ln ∣ x - 3 ∣ + \frac{1}{40} (- ln x^{2} + 2 x + 5 - 4 arctan (\frac{1}{2} (x + 1))) + C

Solution steps

\int \frac{1}{x ^{3} - x ^{2} - x - 15} d x

Take the partial fraction of

\frac{1}{x ^{3} - x ^{2} - x - 15} : \frac{1}{20 ( x - 3 )} + \frac{- x - 5}{20 ( x ^{2} + 2 x + 5 )}

= \int \frac{1}{20 ( x - 3 )} + \frac{- x - 5}{20 ( x ^{2} + 2 x + 5 )} 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 \frac{1}{20 ( x - 3 )} d x + \int \frac{- x - 5}{20 ( x ^{2} + 2 x + 5 )} d x

\int \frac{1}{20 ( x - 3 )} d x = \frac{1}{20} ln ∣ x - 3 ∣

\int \frac{- x - 5}{20 ( x ^{2} + 2 x + 5 )} d x = \frac{1}{40} (- ln x^{2} + 2 x + 5 - 4 arctan (\frac{1}{2} (x + 1)))

= \frac{1}{20} ln ∣ x - 3 ∣ + \frac{1}{40} (- ln x^{2} + 2 x + 5 - 4 arctan (\frac{1}{2} (x + 1)))

Add a constant to the solution

= \frac{1}{20} ln ∣ x - 3 ∣ + \frac{1}{40} (- ln x^{2} + 2 x + 5 - 4 arctan (\frac{1}{2} (x + 1))) + C

\int x^{- 3} ln (- 5 x) d x

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

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

\int \frac{1}{2} tan (x) d x

\int 15 x d x

What is the integral of 1/(x^3-x^2-x-15) ?
The integral of 1/(x^3-x^2-x-15) is 1/20 ln|x-3|+1/40 (-ln|x^2+2x+5|-4arctan(1/2 (x+1)))+C