What is the integral of (28s^3)/(sqrt(8-s^4)) ?

The integral of (28s^3)/(sqrt(8-s^4)) is -14sqrt(8-s^4)+C

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integral of (28s^3)/(sqrt(8-s^4))

\int \frac{28 s ^{3}}{8 - s ^{4}} d s

- 14 8 - s^{4} + C

Solution steps

\int \frac{28 s ^{3}}{8 - s ^{4}} d s

Take the constant out:

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

= 28 \cdot \int \frac{s ^{3}}{8 - s ^{4}} d s

Apply u-substitution

: \int - \frac{1}{4 u} d u

= 28 \cdot \int - \frac{1}{4 u} d u

Take the constant out:

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

= 28 (- \frac{1}{4} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= 28 (- \frac{1}{4} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 8 - s^{4}

= 28 (- \frac{1}{4} \cdot 2 (8 - s^{4})^{\frac{1}{2}})

Simplify

28 (- \frac{1}{4} \cdot 2 (8 - s^{4})^{\frac{1}{2}}) : - 14 8 - s^{4}

= - 14 8 - s^{4}

Add a constant to the solution

= - 14 8 - s^{4} + C

tan (π x)

x \to 0 lim (\frac{4 x}{4 + x - 4 - x})

\int \frac{2 x - 4}{x ^{2} + 4} d x

\int (x^{2} - 25) d x

\frac{\partial}{\partial z} (\frac{x + y}{z - 3 y})

What is the integral of (28s^3)/(sqrt(8-s^4)) ?
The integral of (28s^3)/(sqrt(8-s^4)) is -14sqrt(8-s^4)+C