What is the integral of 3x^2sqrt(2x^3-5) ?

The integral of 3x^2sqrt(2x^3-5) is 1/3 (2x^3-5)^{3/2}+C

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integral of 3x^2sqrt(2x^3-5)

\int 3 x^{2} 2 x^{3} - 5 d x

\frac{1}{3} (2 x^{3} - 5)^{\frac{3}{2}} + C

Solution steps

\int 3 x^{2} 2 x^{3} - 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} 2 x^{3} - 5 d x

Apply u-substitution

= 3 \cdot \int \frac{u}{6} d u

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int u d u

Apply the Power Rule

= 3 \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 x^{3} - 5

= 3 \cdot \frac{1}{6} \cdot \frac{2}{3} (2 x^{3} - 5)^{\frac{3}{2}}

Simplify

3 \cdot \frac{1}{6} \cdot \frac{2}{3} (2 x^{3} - 5)^{\frac{3}{2}} : \frac{1}{3} (2 x^{3} - 5)^{\frac{3}{2}}

= \frac{1}{3} (2 x^{3} - 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (2 x^{3} - 5)^{\frac{3}{2}} + C

t an g e n t f (x) = 4 - x^{2}, at x = 1.154

t an g e n t x^{3} - 3 x^{2} + 2 x - 2, at x = 0

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 2 y})

\frac{d}{d x} (e^{- i x})

x \to 0 lim ((1 + x) \frac{1}{x})

What is the integral of 3x^2sqrt(2x^3-5) ?
The integral of 3x^2sqrt(2x^3-5) is 1/3 (2x^3-5)^{3/2}+C