What is the integral of (2x^2-1)(2x^3-3x+9)^{1/3} ?

The integral of (2x^2-1)(2x^3-3x+9)^{1/3} is 1/4 (2x^3-3x+9)^{4/3}+C

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integral of (2x^2-1)(2x^3-3x+9)^{1/3}

\int (2 x^{2} - 1) (2 x^{3} - 3 x + 9)^{\frac{1}{3}} d x

\frac{1}{4} (2 x^{3} - 3 x + 9)^{\frac{4}{3}} + C

Solution steps

\int (2 x^{2} - 1) (2 x^{3} - 3 x + 9)^{\frac{1}{3}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{3} d u

= \int \frac{u ^{\frac{1}{3}}}{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^{\frac{1}{3}} d u

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

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

Substitute back

u = 2 x^{3} - 3 x + 9

= \frac{1}{3} \cdot \frac{3}{4} (2 x^{3} - 3 x + 9)^{\frac{4}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{4} (2 x^{3} - 3 x + 9)^{\frac{4}{3}} : \frac{1}{4} (2 x^{3} - 3 x + 9)^{\frac{4}{3}}

= \frac{1}{4} (2 x^{3} - 3 x + 9)^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{4} (2 x^{3} - 3 x + 9)^{\frac{4}{3}} + C

\frac{( x ^{2} )}{y ^{2} - 5} y^{^{'}} = \frac{1}{2 y}

\int_{6}^{7} t^{2} d t

x \to 2 lim (f (x))

f (x) = \frac{0.3 ( 9 ^{x} )}{x ^{3}}

y^{^{''}} - 4 y^{^{'}} + 4 y = (12 x^{2} - 6 x) e^{2} x

What is the integral of (2x^2-1)(2x^3-3x+9)^{1/3} ?
The integral of (2x^2-1)(2x^3-3x+9)^{1/3} is 1/4 (2x^3-3x+9)^{4/3}+C