What is the integral of 9x^2\sqrt[4]{7+4x^3} ?

The integral of 9x^2\sqrt[4]{7+4x^3} is 3/5 (4x^3+7)^{5/4}+C

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integral of 9x^2\sqrt[4]{7+4x^3}

\int 9 x^{2} 4 7 + 4 x^{3} d x

\frac{3}{5} (4 x^{3} + 7)^{\frac{5}{4}} + C

Solution steps

\int 9 x^{2} 4 7 + 4 x^{3} d x

Take the constant out:

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

= 9 \cdot \int x^{2} 4 7 + 4 x^{3} d x

Apply u-substitution

: \int \frac{4 u}{12} d u

= 9 \cdot \int \frac{4 u}{12} d u

Take the constant out:

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

= 9 \cdot \frac{1}{12} \cdot \int 4 u d u

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

= 9 \cdot \frac{1}{12} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 7 + 4 x^{3}

= 9 \cdot \frac{1}{12} \cdot \frac{4}{5} (7 + 4 x^{3})^{\frac{5}{4}}

Simplify

9 \cdot \frac{1}{12} \cdot \frac{4}{5} (7 + 4 x^{3})^{\frac{5}{4}} : \frac{3}{5} (4 x^{3} + 7)^{\frac{5}{4}}

= \frac{3}{5} (4 x^{3} + 7)^{\frac{5}{4}}

Add a constant to the solution

= \frac{3}{5} (4 x^{3} + 7)^{\frac{5}{4}} + C

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What is the integral of 9x^2\sqrt[4]{7+4x^3} ?
The integral of 9x^2\sqrt[4]{7+4x^3} is 3/5 (4x^3+7)^{5/4}+C