What is the integral of (y^5)/((1-y^3)^{3/2)} ?

The integral of (y^5)/((1-y^3)^{3/2)} is -(2y^3-4)/(3(-y^3+1)^{1/2)}+C

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integral of (y^5)/((1-y^3)^{3/2)}

\int \frac{y ^{5}}{( 1 - y ^{3} ) ^{\frac{3}{2}}} d y

- \frac{2 y ^{3} - 4}{3 ( - y ^{3} + 1 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{y ^{5}}{( 1 - y ^{3} ) ^{\frac{3}{2}}} d y

Apply u-substitution

: \int \frac{u ^{\frac{1}{5}}}{5 ( - u ^{\frac{3}{5}} + 1 ) ^{\frac{3}{2}}} d u

= \int \frac{u ^{\frac{1}{5}}}{5 ( - u ^{\frac{3}{5}} + 1 ) ^{\frac{3}{2}}} 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 ^{\frac{1}{5}}}{( - u ^{\frac{3}{5}} + 1 ) ^{\frac{3}{2}}} d u

Apply u-substitution

: \int - \frac{5 ( - v + 1 )}{3 v ^{\frac{3}{2}}} d v

= \frac{1}{5} \cdot \int - \frac{5 ( - v + 1 )}{3 v ^{\frac{3}{2}}} d v

Take the constant out:

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

= \frac{1}{5} (- \frac{5}{3} \cdot \int \frac{- v + 1}{v ^{\frac{3}{2}}} d v)

Expand

\frac{- v + 1}{v ^{\frac{3}{2}}} : - \frac{1}{v ^{\frac{1}{2}}} + \frac{1}{v ^{\frac{3}{2}}}

= \frac{1}{5} (- \frac{5}{3} \cdot \int - \frac{1}{v ^{\frac{1}{2}}} + \frac{1}{v ^{\frac{3}{2}}} 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} (- \frac{5}{3} (- \int \frac{1}{v ^{\frac{1}{2}}} d v + \int \frac{1}{v ^{\frac{3}{2}}} d v))

\int \frac{1}{v ^{\frac{1}{2}}} d v = 2 v^{\frac{1}{2}}

\int \frac{1}{v ^{\frac{3}{2}}} d v = - \frac{2}{v ^{\frac{1}{2}}}

= \frac{1}{5} (- \frac{5}{3} (- 2 v^{\frac{1}{2}} - \frac{2}{v ^{\frac{1}{2}}}))

Substitute back

= \frac{1}{5} - \frac{5}{3} - 2 (- (y^{5})^{\frac{3}{5}} + 1)^{\frac{1}{2}} - \frac{2}{( - ( y ^{5} ) ^{\frac{3}{5}} + 1 ) ^{\frac{1}{2}}}

Simplify

\frac{1}{5} - \frac{5}{3} - 2 (- (y^{5})^{\frac{3}{5}} + 1)^{\frac{1}{2}} - \frac{2}{( - ( y ^{5} ) ^{\frac{3}{5}} + 1 ) ^{\frac{1}{2}}} : - \frac{2 y ^{3} - 4}{3 ( - y ^{3} + 1 ) ^{\frac{1}{2}}}

= - \frac{2 y ^{3} - 4}{3 ( - y ^{3} + 1 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{2 y ^{3} - 4}{3 ( - y ^{3} + 1 ) ^{\frac{1}{2}}} + C

f (x) = (x + x^{4}) (- x - 1)

y^{^{'}} sin (2 π x) = π y cos (2 π x)

x \to 1 - lim (4 - x - x^{2})

\frac{4}{s ^{3}}

What is the integral of (y^5)/((1-y^3)^{3/2)} ?
The integral of (y^5)/((1-y^3)^{3/2)} is -(2y^3-4)/(3(-y^3+1)^{1/2)}+C