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

The integral of 3x^2\sqrt[3]{(5x^3+4)^5} is 3/40 (5x^3+4)^{8/3}+C

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integral of 3x^2\sqrt[3]{(5x^3+4)^5}

\int 3 x^{2} 3 (5 x^{3} + 4)^{5} d x

\frac{3}{40} (5 x^{3} + 4)^{\frac{8}{3}} + C

Solution steps

\int 3 x^{2} 3 (5 x^{3} + 4)^{5} d x

Take the constant out:

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

= 3 \cdot \int x^{2} 3 (5 x^{3} + 4)^{5} d x

Apply u-substitution

: \int \frac{u ^{\frac{5}{3}}}{15} d u

= 3 \cdot \int \frac{u ^{\frac{5}{3}}}{15} d u

Take the constant out:

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

= 3 \cdot \frac{1}{15} \cdot \int u^{\frac{5}{3}} d u

Apply the Power Rule

: \frac{3}{8} u^{\frac{8}{3}}

= 3 \cdot \frac{1}{15} \cdot \frac{3}{8} u^{\frac{8}{3}}

Substitute back

u = 5 x^{3} + 4

= 3 \cdot \frac{1}{15} \cdot \frac{3}{8} (5 x^{3} + 4)^{\frac{8}{3}}

Simplify

3 \cdot \frac{1}{15} \cdot \frac{3}{8} (5 x^{3} + 4)^{\frac{8}{3}} : \frac{3}{40} (5 x^{3} + 4)^{\frac{8}{3}}

= \frac{3}{40} (5 x^{3} + 4)^{\frac{8}{3}}

Add a constant to the solution

= \frac{3}{40} (5 x^{3} + 4)^{\frac{8}{3}} + C

\frac{d}{d x} (4 x^{\frac{5}{3}} - 3 x^{- 2} - 12)

2 3 x, at x = 1

\int_{0}^{1} x^{- \frac{1}{2}} d x

\frac{\partial}{\partial x} (\frac{x + 5}{y + 5})

f (θ) = e^{t a n (θ)}

What is the integral of 3x^2\sqrt[3]{(5x^3+4)^5} ?
The integral of 3x^2\sqrt[3]{(5x^3+4)^5} is 3/40 (5x^3+4)^{8/3}+C