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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

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