What is the integral of 6x(4+x^2)^{1/2} ?

The integral of 6x(4+x^2)^{1/2} is 2(x^2+4)^{3/2}+C

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integral of 6x(4+x^2)^{1/2}

\int 6 x (4 + x^{2})^{\frac{1}{2}} d x

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

Solution steps

\int 6 x (4 + x^{2})^{\frac{1}{2}} d x

Take the constant out:

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

= 6 \cdot \int x (4 + x^{2})^{\frac{1}{2}} d x

Apply u-substitution

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

= 6 \cdot \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

= 6 \cdot \frac{1}{2} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 + x^{2}

= 6 \cdot \frac{1}{2} \cdot \frac{2}{3} (4 + x^{2})^{\frac{3}{2}}

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x}, 4 x^{2} + 5 x y - y^{2} = 2

- 3 sin (4 t)

x \cdot \frac{d y}{d x} - x^{2} + x \cdot e^{- x} = 0

\frac{d}{d x} (\frac{x ^{5}}{5})

\frac{d}{d x} (\frac{6}{x + 1})

What is the integral of 6x(4+x^2)^{1/2} ?
The integral of 6x(4+x^2)^{1/2} is 2(x^2+4)^{3/2}+C