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

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

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integral of 1/(x(x^2-4))

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

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

Solution steps

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

Apply u-substitution

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

Take the partial fraction of

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

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

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

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

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

Substitute back

u = x^{2}

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

\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} - 4)

Add a constant to the solution

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

\int x in (x) d x

\int_{0}^{\frac{1}{2}} x 1 - 4 x^{2} d x

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

y^{^{'}} = \frac{x}{200} - \frac{y}{100}

ma c l a u r in 2^{x}

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