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

The integral of 1/((a^3-2a)) is -1/2 ln|a|+1/4 ln|a+sqrt(2)|+1/4 ln|a-sqrt(2)|+C

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integral of 1/((a^3-2a))

integral

\frac{1}{( a ^{3} - 2 a )}

- \frac{1}{2} ln ∣ a ∣ + \frac{1}{4} ln a + 2 + \frac{1}{4} ln a - 2 + C

Solution steps

\int \frac{1}{a ^{3} - 2 a} d a

Take the partial fraction of

\frac{1}{a ^{3} - 2 a} : - \frac{1}{2 a} + \frac{1}{4 ( a + 2 )} + \frac{1}{4 ( a - 2 )}

= \int - \frac{1}{2 a} + \frac{1}{4 ( a + 2 )} + \frac{1}{4 ( a - 2 )} d a

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}{2 a} d a + \int \frac{1}{4 ( a + 2 )} d a + \int \frac{1}{4 ( a - 2 )} d a

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

\int \frac{1}{4 ( a + 2 )} d a = \frac{1}{4} ln a + 2

\int \frac{1}{4 ( a - 2 )} d a = \frac{1}{4} ln a - 2

= - \frac{1}{2} ln ∣ a ∣ + \frac{1}{4} ln a + 2 + \frac{1}{4} ln a - 2

Add a constant to the solution

= - \frac{1}{2} ln ∣ a ∣ + \frac{1}{4} ln a + 2 + \frac{1}{4} ln a - 2 + C

4 - ∣ x ∣

(9 x + 4)^{4}

cos (2 x + 3)

2 x 2 x - 3

\frac{s}{s ^{2} + 9}

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