What is the integral of (32)/(x(x^2-16)) ?

The integral of (32)/(x(x^2-16)) is -2ln(x)+ln|x^2-16|+C

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integral of (32)/(x(x^2-16))

\int \frac{32}{x ( x ^{2} - 16 )} d x

- 2 ln (x) + ln x^{2} - 16 + C

Solution steps

\int \frac{32}{x ( x ^{2} - 16 )} d x

Take the constant out:

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

= 32 \cdot \int \frac{1}{x ( x ^{2} - 16 )} d x

Apply u-substitution

: \int \frac{1}{2 u ( u - 16 )} d u

= 32 \cdot \int \frac{1}{2 u ( u - 16 )} d u

Take the constant out:

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

= 32 \cdot \frac{1}{2} \cdot \int \frac{1}{u ( u - 16 )} d u

Take the partial fraction of

\frac{1}{u ( u - 16 )} : - \frac{1}{16 u} + \frac{1}{16 ( u - 16 )}

= 32 \cdot \frac{1}{2} \cdot \int - \frac{1}{16 u} + \frac{1}{16 ( u - 16 )} 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

= 32 \cdot \frac{1}{2} (- \int \frac{1}{16 u} d u + \int \frac{1}{16 ( u - 16 )} d u)

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

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

= 32 \cdot \frac{1}{2} (- \frac{1}{16} ln ∣ u ∣ + \frac{1}{16} ln ∣ u - 16 ∣)

Substitute back

u = x^{2}

= 32 \cdot \frac{1}{2} (- \frac{1}{16} ln x^{2} + \frac{1}{16} ln x^{2} - 16)

Simplify

32 \cdot \frac{1}{2} (- \frac{1}{16} ln x^{2} + \frac{1}{16} ln x^{2} - 16) : - 2 ln (x) + ln x^{2} - 16

= - 2 ln (x) + ln x^{2} - 16

Add a constant to the solution

= - 2 ln (x) + ln x^{2} - 16 + C

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What is the integral of (32)/(x(x^2-16)) ?
The integral of (32)/(x(x^2-16)) is -2ln(x)+ln|x^2-16|+C