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

The integral of (sqrt(y^5-1))/y is 2/5 (-arctan(sqrt(y^5-1))+sqrt(y^5-1))+C

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

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

\frac{2}{5} (- arctan (y^{5} - 1) + y^{5} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{5 u} d u

= \int \frac{u - 1}{5 u} 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{u - 1}{u} d u

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} + 1} d v

= \frac{1}{5} \cdot \int \frac{2 v ^{2}}{v ^{2} + 1} d v

Take the constant out:

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

= \frac{1}{5} \cdot 2 \cdot \int \frac{v ^{2}}{v ^{2} + 1} d v

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

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

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= \frac{1}{5} \cdot 2 (- arctan (v) + v)

Substitute back

= \frac{1}{5} \cdot 2 (- arctan (y^{5} - 1) + y^{5} - 1)

Simplify

\frac{1}{5} \cdot 2 (- arctan (y^{5} - 1) + y^{5} - 1) : \frac{2}{5} (- arctan (y^{5} - 1) + y^{5} - 1)

= \frac{2}{5} (- arctan (y^{5} - 1) + y^{5} - 1)

Add a constant to the solution

= \frac{2}{5} (- arctan (y^{5} - 1) + y^{5} - 1) + C

\frac{d}{d x} (1 + 3^{s i n (x^{3})})

(x^{3} + 2 x + 6) (4 + \frac{1}{x ^{2}})

\int x^{2} csc (x^{3}) cot (x^{3}) d x

f (x) = - 4 x^{2} - 3 x

\frac{\partial}{\partial x} (x^{2} + y^{2} + 1)

What is the integral of (sqrt(y^5-1))/y ?
The integral of (sqrt(y^5-1))/y is 2/5 (-arctan(sqrt(y^5-1))+sqrt(y^5-1))+C