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

: \int \frac{u}{6} d u

= 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

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

= 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

\frac{d y}{d x} = - k y

x \to 3 lim (∣ x + 1 ∣ + ∣ x - 1 ∣)

arcsec (5 x)

\frac{\partial}{\partial L} (L^{\frac{1}{4}} K^{\frac{1}{4}})

\frac{\partial}{\partial x} ((4 x y - 5)^{2})

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