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

: \int \frac{( u - 4 ) ^{\frac{1}{2}}}{2} d u

= 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

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

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

Apply the Power Rule

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

= 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

\frac{d}{d x} (\frac{e ^{x}}{6 x ^{2} + 1})

\frac{d}{d x} (3000 (1 + \frac{4 x}{x ^{2} + 100}))

\frac{e ^{4 x}}{4} + \frac{e ^{5 x}}{5}

f (x) = 3 x^{2} - 2 cos (x)

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