What is the integral of x^2sqrt(x^3+6) ?

The integral of x^2sqrt(x^3+6) is 2/9 (x^3+6)^{3/2}+C

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integral of x^2sqrt(x^3+6)

\int x^{2} x^{3} + 6 d x

\frac{2}{9} (x^{3} + 6)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x^{3} + 6 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} + 6

= \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 6)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (x^{3} + 6)^{\frac{3}{2}} : \frac{2}{9} (x^{3} + 6)^{\frac{3}{2}}

= \frac{2}{9} (x^{3} + 6)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x^{3} + 6)^{\frac{3}{2}} + C

y = x^{3} - 9 x, (1, - 8)

\int \frac{5 - 4 x ^{2}}{3 + 2 x} d x

y = \frac{e ^{x} + e ^{- x}}{x}

\frac{d}{d x} (\frac{500}{1 + 0.06 x ^{2}})

\int \frac{1}{x + 4} d x

What is the integral of x^2sqrt(x^3+6) ?
The integral of x^2sqrt(x^3+6) is 2/9 (x^3+6)^{3/2}+C