What is the integral of 1/(x(x^{12)+1)^2} ?

The integral of 1/(x(x^{12)+1)^2} is ln(x)-1/12 ln|x^{12}+1|+1/(12(x^{12)+1)}+C

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integral of 1/(x(x^{12)+1)^2}

\int \frac{1}{x ( x ^{12} + 1 ) ^{2}} d x

ln (x) - \frac{1}{12} ln x^{12} + 1 + \frac{1}{12 ( x ^{12} + 1 )} + C

Solution steps

\int \frac{1}{x ( x ^{12} + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{12 u ( u + 1 ) ^{2}} d u

= \int \frac{1}{12 u ( u + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u ( u + 1 ) ^{2}} d u

Apply u-substitution

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

= \frac{1}{12} \cdot \int \frac{1}{( v - 1 ) v ^{2}} d v

Take the partial fraction of

\frac{1}{( v - 1 ) v ^{2}} : \frac{1}{v - 1} - \frac{1}{v} - \frac{1}{v ^{2}}

= \frac{1}{12} \cdot \int \frac{1}{v - 1} - \frac{1}{v} - \frac{1}{v ^{2}} d v

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

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

= \frac{1}{12} (ln ∣ v - 1 ∣ - ln ∣ v ∣ - (- \frac{1}{v}))

Substitute back

= \frac{1}{12} (ln x^{12} + 1 - 1 - ln x^{12} + 1 - (- \frac{1}{x ^{12} + 1}))

Simplify

\frac{1}{12} (ln x^{12} + 1 - 1 - ln x^{12} + 1 - (- \frac{1}{x ^{12} + 1})) : ln (x) - \frac{1}{12} ln x^{12} + 1 + \frac{1}{12 ( x ^{12} + 1 )}

= ln (x) - \frac{1}{12} ln x^{12} + 1 + \frac{1}{12 ( x ^{12} + 1 )}

Add a constant to the solution

= ln (x) - \frac{1}{12} ln x^{12} + 1 + \frac{1}{12 ( x ^{12} + 1 )} + C

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What is the integral of 1/(x(x^{12)+1)^2} ?
The integral of 1/(x(x^{12)+1)^2} is ln(x)-1/12 ln|x^{12}+1|+1/(12(x^{12)+1)}+C