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

The integral of 4x^2\sqrt[4]{x^3+7} is 16/15 (x^3+7)^{5/4}+C

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

\frac{16}{15} (x^{3} + 7)^{\frac{5}{4}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

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

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

Substitute back

u = x^{3} + 7

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

Simplify

4 \cdot \frac{1}{3} \cdot \frac{4}{5} (x^{3} + 7)^{\frac{5}{4}} : \frac{16}{15} (x^{3} + 7)^{\frac{5}{4}}

= \frac{16}{15} (x^{3} + 7)^{\frac{5}{4}}

Add a constant to the solution

= \frac{16}{15} (x^{3} + 7)^{\frac{5}{4}} + C

\int - 2 x^{2} e^{- x} d x

\int \frac{5}{x ^{2} + 6} d x

cos^{2} (θ)

\int \frac{0 ^{1} ( x ^{2} - 2 x - 1 )}{x} d x

\int (\frac{2}{x + 1}) d x

What is the integral of 4x^2\sqrt[4]{x^3+7} ?
The integral of 4x^2\sqrt[4]{x^3+7} is 16/15 (x^3+7)^{5/4}+C