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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int \frac{2 u ^{2}}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int u^{2} d u

Apply the Power Rule

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

Substitute back

u = 3 + x^{3}

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

Simplify

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

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

Add a constant to the solution

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

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