What is the integral of (15x(4-x^2)^{1/4})/2 ?

The integral of (15x(4-x^2)^{1/4})/2 is -3(-x^2+4)^{5/4}+C

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integral of (15x(4-x^2)^{1/4})/2

\int \frac{15 x ( 4 - x ^{2} ) ^{\frac{1}{4}}}{2} d x

- 3 (- x^{2} + 4)^{\frac{5}{4}} + C

Solution steps

\int \frac{15 x ( 4 - x ^{2} ) ^{\frac{1}{4}}}{2} d x

Take the constant out:

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

= \frac{15}{2} \cdot \int x (4 - x^{2})^{\frac{1}{4}} d x

Apply u-substitution

= \frac{15}{2} \cdot \int \frac{( 4 - u ) ^{\frac{1}{4}}}{2} d u

Take the constant out:

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

= \frac{15}{2} \cdot \frac{1}{2} \cdot \int (4 - u)^{\frac{1}{4}} d u

Apply u-substitution

= \frac{15}{2} \cdot \frac{1}{2} \cdot \int - v^{\frac{1}{4}} d v

Take the constant out:

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

= \frac{15}{2} \cdot \frac{1}{2} (- \int v^{\frac{1}{4}} d v)

Apply the Power Rule

= \frac{15}{2} \cdot \frac{1}{2} (- \frac{4}{5} v^{\frac{5}{4}})

Substitute back

= \frac{15}{2} \cdot \frac{1}{2} (- \frac{4}{5} (4 - x^{2})^{\frac{5}{4}})

Simplify

\frac{15}{2} \cdot \frac{1}{2} (- \frac{4}{5} (4 - x^{2})^{\frac{5}{4}}) : - 3 (- x^{2} + 4)^{\frac{5}{4}}

= - 3 (- x^{2} + 4)^{\frac{5}{4}}

Add a constant to the solution

= - 3 (- x^{2} + 4)^{\frac{5}{4}} + C

What is the integral of (15x(4-x^2)^{1/4})/2 ?
The integral of (15x(4-x^2)^{1/4})/2 is -3(-x^2+4)^{5/4}+C