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Popular >

integral of 1/(x^3-64)

Solution

\int \frac{1}{x ^{3} - 64} d x

Solution

\frac{1}{48} ln ∣ x - 4 ∣ - \frac{1}{96} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3}) + C

Solution steps

\int \frac{1}{x ^{3} - 64} d x

Take the partial fraction of

\frac{1}{x ^{3} - 64} : \frac{1}{48 ( x - 4 )} + \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )}

= \int \frac{1}{48 ( x - 4 )} + \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )} 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}{48 ( x - 4 )} d x + \int \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )} d x

\int \frac{1}{48 ( x - 4 )} d x = \frac{1}{48} ln ∣ x - 4 ∣

\int \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )} d x = \frac{1}{96 3} (- 3 ln \frac{( x + 2 ) ^{2}}{12} + 1 - 6 arctan (\frac{1}{2 3} x + \frac{1}{3}))

= \frac{1}{48} ln ∣ x - 4 ∣ + \frac{1}{96 3} (- 3 ln \frac{( x + 2 ) ^{2}}{12} + 1 - 6 arctan (\frac{1}{2 3} x + \frac{1}{3}))

Simplify

\frac{1}{48} ln ∣ x - 4 ∣ + \frac{1}{96 3} (- 3 ln \frac{( x + 2 ) ^{2}}{12} + 1 - 6 arctan (\frac{1}{2 3} x + \frac{1}{3})) : \frac{1}{48} ln ∣ x - 4 ∣ - \frac{1}{96} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3})

= \frac{1}{48} ln ∣ x - 4 ∣ - \frac{1}{96} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3})

Add a constant to the solution

= \frac{1}{48} ln ∣ x - 4 ∣ - \frac{1}{96} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3}) + C

Graph

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Frequently Asked Questions (FAQ)

What is the integral of 1/(x^3-64) ?
The integral of 1/(x^3-64) is