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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ( u - 2 ) ^{2}} d u

Take the partial fraction of

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

= \frac{1}{2} \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u - 2 )} + \frac{1}{2 ( u - 2 ) ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

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

= \frac{1}{2} (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u - 2 ∣ - \frac{1}{2 ( u - 2 )})

Substitute back

u = x^{2}

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

\frac{1}{4} ln x^{2} = \frac{1}{2} ln (x)

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

Add a constant to the solution

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

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