What is the integral of x^2sqrt(x^3+39) ?

The integral of x^2sqrt(x^3+39) is 2/9 (x^3+39)^{3/2}+C

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integral of x^2sqrt(x^3+39)

\int x^{2} x^{3} + 39 d x

\frac{2}{9} (x^{3} + 39)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x^{3} + 39 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} + 39

= \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 39)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (x^{3} + 39)^{\frac{3}{2}} : \frac{2}{9} (x^{3} + 39)^{\frac{3}{2}}

= \frac{2}{9} (x^{3} + 39)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x^{3} + 39)^{\frac{3}{2}} + C

\int \frac{x}{( 1 - x ) ^{2}} d x

x \to 0 lim (\frac{9 x ^{2} - 8 x}{x})

- y + y^{^{'}} = x + e^{2} x

y^{^{'}} + \frac{2}{x} y = x^{2} y^{2}

\frac{d}{d x} (\frac{cos ( x )}{x})

What is the integral of x^2sqrt(x^3+39) ?
The integral of x^2sqrt(x^3+39) is 2/9 (x^3+39)^{3/2}+C