What is the integral of x(x^2-36)^{1/2} ?

The integral of x(x^2-36)^{1/2} is 1/3 (x^2-36)^{3/2}+C

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integral of x(x^2-36)^{1/2}

\int x (x^{2} - 36)^{\frac{1}{2}} d x

\frac{1}{3} (x^{2} - 36)^{\frac{3}{2}} + C

Solution steps

\int x (x^{2} - 36)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{2} d u

= \int \frac{u ^{\frac{1}{2}}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{2} - 36

= \frac{1}{2} \cdot \frac{2}{3} (x^{2} - 36)^{\frac{3}{2}}

Simplify

\frac{1}{2} \cdot \frac{2}{3} (x^{2} - 36)^{\frac{3}{2}} : \frac{1}{3} (x^{2} - 36)^{\frac{3}{2}}

= \frac{1}{3} (x^{2} - 36)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (x^{2} - 36)^{\frac{3}{2}} + C

\frac{1}{( 3 - x ) ^{2}} + \frac{1}{x}, 2

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

y = x^{2} + 4, (- 2, 8)

f (x) = 3 x^{\frac{2}{3}} - 2 x

x \to 0 - lim (- \frac{1}{x})

What is the integral of x(x^2-36)^{1/2} ?
The integral of x(x^2-36)^{1/2} is 1/3 (x^2-36)^{3/2}+C