What is the integral of (x^3+3)^{1/4}x^5 ?

The integral of (x^3+3)^{1/4}x^5 is 1/3 (4/9 (x^3+3)^{9/4}-12/5 (x^3+3)^{5/4})+C

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integral of (x^3+3)^{1/4}x^5

\int (x^{3} + 3)^{\frac{1}{4}} x^{5} d x

\frac{1}{3} (\frac{4}{9} (x^{3} + 3)^{\frac{9}{4}} - \frac{12}{5} (x^{3} + 3)^{\frac{5}{4}}) + C

Solution steps

\int (x^{3} + 3)^{\frac{1}{4}} x^{5} d x

Apply u-substitution

: \int \frac{( u - 3 ) u ^{\frac{1}{4}}}{3} d u

= \int \frac{( u - 3 ) u ^{\frac{1}{4}}}{3} 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 (u - 3) u^{\frac{1}{4}} d u

Expand

(u - 3) u^{\frac{1}{4}} : u^{\frac{5}{4}} - 3 u^{\frac{1}{4}}

= \frac{1}{3} \cdot \int u^{\frac{5}{4}} - 3 u^{\frac{1}{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

= \frac{1}{3} (\int u^{\frac{5}{4}} d u - \int 3 u^{\frac{1}{4}} d u)

\int u^{\frac{5}{4}} d u = \frac{4}{9} u^{\frac{9}{4}}

\int 3 u^{\frac{1}{4}} d u = \frac{12}{5} u^{\frac{5}{4}}

= \frac{1}{3} (\frac{4}{9} u^{\frac{9}{4}} - \frac{12}{5} u^{\frac{5}{4}})

Substitute back

u = x^{3} + 3

= \frac{1}{3} (\frac{4}{9} (x^{3} + 3)^{\frac{9}{4}} - \frac{12}{5} (x^{3} + 3)^{\frac{5}{4}})

Add a constant to the solution

= \frac{1}{3} (\frac{4}{9} (x^{3} + 3)^{\frac{9}{4}} - \frac{12}{5} (x^{3} + 3)^{\frac{5}{4}}) + C

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What is the integral of (x^3+3)^{1/4}x^5 ?
The integral of (x^3+3)^{1/4}x^5 is 1/3 (4/9 (x^3+3)^{9/4}-12/5 (x^3+3)^{5/4})+C