What is the integral of x^2sqrt(4+x^3) ?

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

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integral of x^2sqrt(4+x^3)

\int x^{2} 4 + x^{3} d x

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

Solution steps

\int x^{2} 4 + x^{3} d x

Apply u-substitution

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

= \int \frac{u}{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 d u

Apply the Power Rule

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

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

Substitute back

u = 4 + x^{3}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (2 x^{2} + 4)

y^{^{'}} + 5 y = 3

\frac{x ^{4}}{4} + \frac{7}{6} x^{3} - x^{2} + 6 x - 8

x \cdot 3^{x^{2} + 1}

6 (t + 1) (\frac{d y}{d t}) - 5 y = 5 t

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