What is the integral of 1/(x(x^5+1)^2) ?

The integral of 1/(x(x^5+1)^2) is 1/5 (ln|x^5|-ln|x^5+1|+1/(x^5+1))+C

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integral of 1/(x(x^5+1)^2)

\int \frac{1}{x ( x ^{5} + 1 ) ^{2}} d x

\frac{1}{5} (ln x^{5} - ln x^{5} + 1 + \frac{1}{x ^{5} + 1}) + C

Solution steps

\int \frac{1}{x ( x ^{5} + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{5 u ( u + 1 ) ^{2}} d u

= \int \frac{1}{5 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}{5} \cdot \int \frac{1}{u ( u + 1 ) ^{2}} d u

Apply u-substitution

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

= \frac{1}{5} \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}{5} \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}{5} (\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}{5} (ln ∣ v - 1 ∣ - ln ∣ v ∣ - (- \frac{1}{v}))

Substitute back

= \frac{1}{5} (ln x^{5} + 1 - 1 - ln x^{5} + 1 - (- \frac{1}{x ^{5} + 1}))

Simplify

= \frac{1}{5} (ln x^{5} - ln x^{5} + 1 + \frac{1}{x ^{5} + 1})

Add a constant to the solution

= \frac{1}{5} (ln x^{5} - ln x^{5} + 1 + \frac{1}{x ^{5} + 1}) + C

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What is the integral of 1/(x(x^5+1)^2) ?
The integral of 1/(x(x^5+1)^2) is 1/5 (ln|x^5|-ln|x^5+1|+1/(x^5+1))+C