What is the integral of x\sqrt[4]{x+20} ?

The integral of x\sqrt[4]{x+20} is 4(1/9 (x+20)^{9/4}-4(x+20)^{5/4})+C

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integral of x\sqrt[4]{x+20}

4 (\frac{1}{9} (x + 20)^{\frac{9}{4}} - 4 (x + 20)^{\frac{5}{4}}) + C

Solution steps

Apply u-substitution

: \int 4 u^{4} (u^{4} - 20) d u

= \int 4 u^{4} (u^{4} - 20) d u

Take the constant out:

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

= 4 \cdot \int u^{4} (u^{4} - 20) d u

Expand

u^{4} (u^{4} - 20) : u^{8} - 20 u^{4}

= 4 \cdot \int u^{8} - 20 u^{4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int u^{8} d u - \int 20 u^{4} d u)

\int u^{8} d u = \frac{u ^{9}}{9}

\int 20 u^{4} d u = 4 u^{5}

= 4 (\frac{u ^{9}}{9} - 4 u^{5})

Substitute back

Simplify

= 4 (\frac{1}{9} (x + 20)^{\frac{9}{4}} - 4 (x + 20)^{\frac{5}{4}})

Add a constant to the solution

= 4 (\frac{1}{9} (x + 20)^{\frac{9}{4}} - 4 (x + 20)^{\frac{5}{4}}) + C

\int \frac{1}{x + 1} d x

\frac{\partial}{\partial y} (x^{2} - 4 x y + y^{3} + 4 y)

\int (2 x yz + sin (x)) d x

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

\frac{d}{d x} (- \frac{5}{( x + 3 ) ^{3}})

What is the integral of x\sqrt[4]{x+20} ?
The integral of x\sqrt[4]{x+20} is 4(1/9 (x+20)^{9/4}-4(x+20)^{5/4})+C