What is the integral of 2/(x(x^2+6)^2) ?

The integral of 2/(x(x^2+6)^2) is 1/18 ln(x)-1/36 ln|x^2+6|+1/(6(x^2+6))+C

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integral of 2/(x(x^2+6)^2)

\int \frac{2}{x ( x ^{2} + 6 ) ^{2}} d x

\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )} + C

Solution steps

\int \frac{2}{x ( x ^{2} + 6 ) ^{2}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ( x ^{2} + 6 ) ^{2}} d x

Apply u-substitution

= 2 \cdot \int \frac{1}{2 ( u - 6 ) u ^{2}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 6 ) u ^{2}} d u

Take the partial fraction of

\frac{1}{( u - 6 ) u ^{2}} : \frac{1}{36 ( u - 6 )} - \frac{1}{36 u} - \frac{1}{6 u ^{2}}

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{36 ( u - 6 )} - \frac{1}{36 u} - \frac{1}{6 u ^{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

= 2 \cdot \frac{1}{2} (\int \frac{1}{36 ( u - 6 )} d u - \int \frac{1}{36 u} d u - \int \frac{1}{6 u ^{2}} d u)

\int \frac{1}{36 ( u - 6 )} d u = \frac{1}{36} ln ∣ u - 6 ∣

\int \frac{1}{36 u} d u = \frac{1}{36} ln ∣ u ∣

\int \frac{1}{6 u ^{2}} d u = - \frac{1}{6 u}

= 2 \cdot \frac{1}{2} (\frac{1}{36} ln ∣ u - 6 ∣ - \frac{1}{36} ln ∣ u ∣ - (- \frac{1}{6 u}))

Substitute back

u = x^{2} + 6

= 2 \cdot \frac{1}{2} (\frac{1}{36} ln x^{2} + 6 - 6 - \frac{1}{36} ln x^{2} + 6 - (- \frac{1}{6 ( x ^{2} + 6 )}))

Simplify

2 \cdot \frac{1}{2} (\frac{1}{36} ln x^{2} + 6 - 6 - \frac{1}{36} ln x^{2} + 6 - (- \frac{1}{6 ( x ^{2} + 6 )})) : \frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )}

= \frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )}

Add a constant to the solution

= \frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )} + C

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What is the integral of 2/(x(x^2+6)^2) ?
The integral of 2/(x(x^2+6)^2) is 1/18 ln(x)-1/36 ln|x^2+6|+1/(6(x^2+6))+C