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

The integral of x(x+6)^{1/2} is 2(1/5 (x+6)^{5/2}-2(x+6)^{3/2})+C

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

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

2 (\frac{1}{5} (x + 6)^{\frac{5}{2}} - 2 (x + 6)^{\frac{3}{2}}) + C

Solution steps

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

Apply u-substitution

: \int 2 (u^{2} - 6) u^{2} d u

= \int 2 (u^{2} - 6) u^{2} d u

Take the constant out:

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

= 2 \cdot \int (u^{2} - 6) u^{2} d u

Expand

(u^{2} - 6) u^{2} : u^{4} - 6 u^{2}

= 2 \cdot \int u^{4} - 6 u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int u^{4} d u - \int 6 u^{2} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 6 u^{2} d u = 2 u^{3}

= 2 (\frac{u ^{5}}{5} - 2 u^{3})

Substitute back

u = (x + 6)^{\frac{1}{2}}

= 2 \frac{( ( x + 6 ) ^{\frac{1}{2}} ) ^{5}}{5} - 2 ((x + 6)^{\frac{1}{2}})^{3}

Simplify

2 \frac{( ( x + 6 ) ^{\frac{1}{2}} ) ^{5}}{5} - 2 ((x + 6)^{\frac{1}{2}})^{3} : 2 (\frac{1}{5} (x + 6)^{\frac{5}{2}} - 2 (x + 6)^{\frac{3}{2}})

= 2 (\frac{1}{5} (x + 6)^{\frac{5}{2}} - 2 (x + 6)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{1}{5} (x + 6)^{\frac{5}{2}} - 2 (x + 6)^{\frac{3}{2}}) + C

\int \frac{1}{5 x + 2} d x

\int_{1}^{2} x e^{3 x^{2}} d x

\frac{d y}{d x} = \frac{y}{x}

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

(4 + x) y^{^{'}} = y^{3}

What is the integral of x(x+6)^{1/2} ?
The integral of x(x+6)^{1/2} is 2(1/5 (x+6)^{5/2}-2(x+6)^{3/2})+C