What is the integral of 3x(x^2-4)^{1/2} ?

The integral of 3x(x^2-4)^{1/2} is (x^2-4)^{3/2}+C

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integral of 3x(x^2-4)^{1/2}

\int 3 x (x^{2} - 4)^{\frac{1}{2}} d x

(x^{2} - 4)^{\frac{3}{2}} + C

Solution steps

\int 3 x (x^{2} - 4)^{\frac{1}{2}} d x

Take the constant out:

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

= 3 \cdot \int x (x^{2} - 4)^{\frac{1}{2}} d x

Apply u-substitution

= 3 \cdot \int \frac{( u - 4 ) ^{\frac{1}{2}}}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int (u - 4)^{\frac{1}{2}} d u

Apply u-substitution

= 3 \cdot \frac{1}{2} \cdot \int v^{\frac{1}{2}} d v

Apply the Power Rule

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

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{2}{3} (x^{2} - 4)^{\frac{3}{2}}

Simplify

3 \cdot \frac{1}{2} \cdot \frac{2}{3} (x^{2} - 4)^{\frac{3}{2}} : (x^{2} - 4)^{\frac{3}{2}}

= (x^{2} - 4)^{\frac{3}{2}}

Add a constant to the solution

= (x^{2} - 4)^{\frac{3}{2}} + C

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What is the integral of 3x(x^2-4)^{1/2} ?
The integral of 3x(x^2-4)^{1/2} is (x^2-4)^{3/2}+C