What is the integral of (y^5)/(1-y^4) ?

The integral of (y^5)/(1-y^4) is -(y^2)/2+1/4 (ln|y^2+1|-ln|y^2-1|)+C

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integral of (y^5)/(1-y^4)

\int \frac{y ^{5}}{1 - y ^{4}} d y

- \frac{y ^{2}}{2} + \frac{1}{4} (ln y^{2} + 1 - ln y^{2} - 1) + C

Solution steps

\int \frac{y ^{5}}{1 - y ^{4}} d y

Long division

\frac{y ^{5}}{1 - y ^{4}} : - y + \frac{y}{- y ^{4} + 1}

= \int - y + \frac{y}{- y ^{4} + 1} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int y d y + \int \frac{y}{- y ^{4} + 1} d y

\int y d y = \frac{y ^{2}}{2}

\int \frac{y}{- y ^{4} + 1} d y = \frac{1}{4} (ln y^{2} + 1 - ln y^{2} - 1)

= - \frac{y ^{2}}{2} + \frac{1}{4} (ln y^{2} + 1 - ln y^{2} - 1)

Add a constant to the solution

= - \frac{y ^{2}}{2} + \frac{1}{4} (ln y^{2} + 1 - ln y^{2} - 1) + C

\frac{\partial}{\partial t} (\frac{s}{t})

x \to 4 lim (2 x^{4} - x^{2} - 8 x)

\frac{\partial}{\partial x} (\frac{3}{3 x + 5 y})

6 x^{2} ln (x), 24 ln (x)

What is the integral of (y^5)/(1-y^4) ?
The integral of (y^5)/(1-y^4) is -(y^2)/2+1/4 (ln|y^2+1|-ln|y^2-1|)+C