What is the integral of sqrt((x^4)/(x^3-1)) ?

The integral of sqrt((x^4)/(x^3-1)) is 2/3 sqrt(x^3-1)+C

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integral of sqrt((x^4)/(x^3-1))

\int \frac{x ^{4}}{x ^{3} - 1} d x

\frac{2}{3} x^{3} - 1 + C

Solution steps

\int \frac{x ^{4}}{x ^{3} - 1} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 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} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{3} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{3} - 1

= \frac{1}{3} \cdot 2 (x^{3} - 1)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot 2 (x^{3} - 1)^{\frac{1}{2}} : \frac{2}{3} x^{3} - 1

= \frac{2}{3} x^{3} - 1

Add a constant to the solution

= \frac{2}{3} x^{3} - 1 + C

y = \frac{4}{4 - x ^{2}}, (0, 1)

t y^{^{'}} + y = 8

\frac{2 t}{t ^{2} + 4}

(5, - 3), (0, - 9)

\frac{\partial}{\partial y} (e^{3 x} \cdot tan (y))

What is the integral of sqrt((x^4)/(x^3-1)) ?
The integral of sqrt((x^4)/(x^3-1)) is 2/3 sqrt(x^3-1)+C