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

The integral of sqrt(1+6x^{1/2)} is 2/135 (9sqrt(x)-1)(6sqrt(x)+1)^{3/2}+C

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

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

\frac{2}{135} (9 x - 1) (6 x + 1)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

= \int 1 + 6 u \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int 1 + 6 u u d u

Apply u-substitution

= 2 \cdot \int \frac{v ( v - 1 )}{36} d v

Take the constant out:

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

= 2 \cdot \frac{1}{36} \cdot \int v (v - 1) d v

Expand

v (v - 1) : v^{\frac{3}{2}} - v

= 2 \cdot \frac{1}{36} \cdot \int v^{\frac{3}{2}} - v d v

Apply the Sum Rule:

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

= 2 \cdot \frac{1}{36} (\int v^{\frac{3}{2}} d v - \int v d v)

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

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

= 2 \cdot \frac{1}{36} (\frac{2}{5} v^{\frac{5}{2}} - \frac{2}{3} v^{\frac{3}{2}})

Substitute back

= 2 \cdot \frac{1}{36} (\frac{2}{5} (1 + 6 x^{\frac{1}{2}})^{\frac{5}{2}} - \frac{2}{3} (1 + 6 x^{\frac{1}{2}})^{\frac{3}{2}})

Simplify

2 \cdot \frac{1}{36} (\frac{2}{5} (1 + 6 x^{\frac{1}{2}})^{\frac{5}{2}} - \frac{2}{3} (1 + 6 x^{\frac{1}{2}})^{\frac{3}{2}}) : \frac{2}{135} (9 x - 1) (6 x + 1)^{\frac{3}{2}}

= \frac{2}{135} (9 x - 1) (6 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{135} (9 x - 1) (6 x + 1)^{\frac{3}{2}} + C

What is the integral of sqrt(1+6x^{1/2)} ?
The integral of sqrt(1+6x^{1/2)} is 2/135 (9sqrt(x)-1)(6sqrt(x)+1)^{3/2}+C