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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = x^{3} - 2

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} + 3 x y - 3 = 0

x \to 1 lim (3 x^{2} - 7 x + 1)

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

- sin (x) + cos (x) + x - \frac{1}{x ^{2}}

\frac{d}{d x} (x^{3} - 6 x^{2} + 8 x)

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