What is the integral of 1/(y(y^3-1)) ?

The integral of 1/(y(y^3-1)) is 1/3 (-ln|y^3|+ln|-1+y^3|)+C

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integral of 1/(y(y^3-1))

\int \frac{1}{y ( y ^{3} - 1 )} d y

\frac{1}{3} (- ln y^{3} + ln - 1 + y^{3}) + C

Solution steps

\int \frac{1}{y ( y ^{3} - 1 )} d y

Apply u-substitution

: \int \frac{1}{3 ( u + 1 ) u} d u

= \int \frac{1}{3 ( u + 1 ) u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{( u + 1 ) u} d u

Take the partial fraction of

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

= \frac{1}{3} \cdot \int - \frac{1}{u + 1} + \frac{1}{u} 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}{3} (- \int \frac{1}{u + 1} d u + \int \frac{1}{u} d u)

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

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{3} (- ln ∣ u + 1 ∣ + ln ∣ u ∣)

Substitute back

u = - 1 + y^{3}

= \frac{1}{3} (- ln - 1 + y^{3} + 1 + ln - 1 + y^{3})

Simplify

\frac{1}{3} (- ln - 1 + y^{3} + 1 + ln - 1 + y^{3}) : \frac{1}{3} (- ln y^{3} + ln - 1 + y^{3})

= \frac{1}{3} (- ln y^{3} + ln - 1 + y^{3})

Add a constant to the solution

= \frac{1}{3} (- ln y^{3} + ln - 1 + y^{3}) + C

y^{^{''}} - 6 y^{^{'}} = 4 x

\int \frac{24}{( 399 x - 1 ) ( 399 + 1 )} d x

arcsin (u)

\frac{d}{d x} (3 x^{2} + 2 x y - y^{2})

f (x) = (3 - x)^{4} (x + 5)^{3}

What is the integral of 1/(y(y^3-1)) ?
The integral of 1/(y(y^3-1)) is 1/3 (-ln|y^3|+ln|-1+y^3|)+C