What is the integral of x^{5/3}sqrt(x^{4/3)+9} ?

The integral of x^{5/3}sqrt(x^{4/3)+9} is 3/10 (x^{4/3}-6)(x^{4/3}+9)^{3/2}+C

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integral of x^{5/3}sqrt(x^{4/3)+9}

\int x^{\frac{5}{3}} x^{\frac{4}{3}} + 9 d x

\frac{3}{10} (x^{\frac{4}{3}} - 6) (x^{\frac{4}{3}} + 9)^{\frac{3}{2}} + C

Solution steps

\int x^{\frac{5}{3}} x^{\frac{4}{3}} + 9 d x

Apply u-substitution

= \int \frac{3 ( u - 9 ) u ^{\frac{1}{2}}}{4} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int (u - 9) u^{\frac{1}{2}} d u

Expand

(u - 9) u^{\frac{1}{2}} : u^{\frac{3}{2}} - 9 u^{\frac{1}{2}}

= \frac{3}{4} \cdot \int u^{\frac{3}{2}} - 9 u^{\frac{1}{2}} 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{3}{4} (\int u^{\frac{3}{2}} d u - \int 9 u^{\frac{1}{2}} d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 9 u^{\frac{1}{2}} d u = 6 u^{\frac{3}{2}}

= \frac{3}{4} (\frac{2}{5} u^{\frac{5}{2}} - 6 u^{\frac{3}{2}})

Substitute back

u = x^{\frac{4}{3}} + 9

= \frac{3}{4} (\frac{2}{5} (x^{\frac{4}{3}} + 9)^{\frac{5}{2}} - 6 (x^{\frac{4}{3}} + 9)^{\frac{3}{2}})

Simplify

\frac{3}{4} (\frac{2}{5} (x^{\frac{4}{3}} + 9)^{\frac{5}{2}} - 6 (x^{\frac{4}{3}} + 9)^{\frac{3}{2}}) : \frac{3}{10} (x^{\frac{4}{3}} - 6) (x^{\frac{4}{3}} + 9)^{\frac{3}{2}}

= \frac{3}{10} (x^{\frac{4}{3}} - 6) (x^{\frac{4}{3}} + 9)^{\frac{3}{2}}

Add a constant to the solution

= \frac{3}{10} (x^{\frac{4}{3}} - 6) (x^{\frac{4}{3}} + 9)^{\frac{3}{2}} + C

What is the integral of x^{5/3}sqrt(x^{4/3)+9} ?
The integral of x^{5/3}sqrt(x^{4/3)+9} is 3/10 (x^{4/3}-6)(x^{4/3}+9)^{3/2}+C