What is the integral of x^2sqrt(x^3+37) ?

The integral of x^2sqrt(x^3+37) is 2/9 (x^3+37)^{3/2}+C

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integral of x^2sqrt(x^3+37)

\int x^{2} x^{3} + 37 d x

\frac{2}{9} (x^{3} + 37)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x^{3} + 37 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} + 37

= \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 37)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (x^{3} + 37)^{\frac{3}{2}} : \frac{2}{9} (x^{3} + 37)^{\frac{3}{2}}

= \frac{2}{9} (x^{3} + 37)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x^{3} + 37)^{\frac{3}{2}} + C

\int \frac{x + 2}{x - 1} d x

f (x) = 7 x + 2, at x = 2

(0, 1), (5, 4)

(21 + x^{2}) y^{^{'}} - 2 x y = 0

\frac{\partial}{\partial y} (x y^{3} + x^{2} y^{2})

What is the integral of x^2sqrt(x^3+37) ?
The integral of x^2sqrt(x^3+37) is 2/9 (x^3+37)^{3/2}+C