What is the integral of x^2sqrt(x^3+35) ?

The integral of x^2sqrt(x^3+35) is 2/9 (x^3+35)^{3/2}+C

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integral of x^2sqrt(x^3+35)

\int x^{2} x^{3} + 35 d x

\frac{2}{9} (x^{3} + 35)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x^{3} + 35 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} + 35

= \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 35)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (x^{3} + 35)^{\frac{3}{2}} : \frac{2}{9} (x^{3} + 35)^{\frac{3}{2}}

= \frac{2}{9} (x^{3} + 35)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x^{3} + 35)^{\frac{3}{2}} + C

\frac{d}{d x} (1 + 1 - x)

y^{^{'}} - 5 y = 2 e^{6 t}

y = 2 x, y = x^{2} - 4 x

\frac{\partial}{\partial x} (- 7 x^{9})

y^{^{'}} = x sin (x) + x cos (x)

What is the integral of x^2sqrt(x^3+35) ?
The integral of x^2sqrt(x^3+35) is 2/9 (x^3+35)^{3/2}+C