What is the integral of x^2sqrt(x^3+19) ?

The integral of x^2sqrt(x^3+19) is 2/9 (x^3+19)^{3/2}+C

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integral of x^2sqrt(x^3+19)

\int x^{2} x^{3} + 19 d x

\frac{2}{9} (x^{3} + 19)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x^{3} + 19 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 = x^{3} + 19

= \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 19)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (x^{3} + 19)^{\frac{3}{2}} : \frac{2}{9} (x^{3} + 19)^{\frac{3}{2}}

= \frac{2}{9} (x^{3} + 19)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x^{3} + 19)^{\frac{3}{2}} + C

\frac{d y}{d x} - \frac{2}{x} y = \frac{3}{x ^{2}} y^{4}

\int_{1}^{2} 13 4 x^{2} - 3 d x

\frac{d}{d x} (- 5 x + 2)

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

- \frac{5}{4 y ^{3}}

What is the integral of x^2sqrt(x^3+19) ?
The integral of x^2sqrt(x^3+19) is 2/9 (x^3+19)^{3/2}+C