What is the integral of 1/(x^3-2x^2-4x+8) ?

The integral of 1/(x^3-2x^2-4x+8) is -1/16 ln|x-2|-1/(4(x-2))+1/16 ln|x+2|+C

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integral of 1/(x^3-2x^2-4x+8)

\int \frac{1}{x ^{3} - 2 x ^{2} - 4 x + 8} d x

- \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )} + \frac{1}{16} ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{1}{x ^{3} - 2 x ^{2} - 4 x + 8} d x

Take the partial fraction of

\frac{1}{x ^{3} - 2 x ^{2} - 4 x + 8} : - \frac{1}{16 ( x - 2 )} + \frac{1}{4 ( x - 2 ) ^{2}} + \frac{1}{16 ( x + 2 )}

= \int - \frac{1}{16 ( x - 2 )} + \frac{1}{4 ( x - 2 ) ^{2}} + \frac{1}{16 ( 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 \frac{1}{16 ( x - 2 )} d x + \int \frac{1}{4 ( x - 2 ) ^{2}} d x + \int \frac{1}{16 ( x + 2 )} d x

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

\int \frac{1}{4 ( x - 2 ) ^{2}} d x = - \frac{1}{4 ( x - 2 )}

\int \frac{1}{16 ( x + 2 )} d x = \frac{1}{16} ln ∣ x + 2 ∣

= - \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )} + \frac{1}{16} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )} + \frac{1}{16} ln ∣ x + 2 ∣ + C

\int_{0}^{1} (6 x^{2} - 1) d x

f (x) = x^{2} - 3

\frac{\partial}{\partial x} (y^{2} e^{- 2 x})

\int 10 x - 7 d x

What is the integral of 1/(x^3-2x^2-4x+8) ?
The integral of 1/(x^3-2x^2-4x+8) is -1/16 ln|x-2|-1/(4(x-2))+1/16 ln|x+2|+C