What is the integral of sqrt(x)sqrt(1+1/(4x)) ?

The integral of sqrt(x)sqrt(1+1/(4x)) is 1/12 (4x+1)^{3/2}+C

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integral of sqrt(x)sqrt(1+1/(4x))

\int x 1 + \frac{1}{4 x} d x

\frac{1}{12} (4 x + 1)^{\frac{3}{2}} + C

Solution steps

\int x 1 + \frac{1}{4 x} d x

Simplify

x 1 + \frac{1}{4 x} : \frac{4 x + 1}{2}

= \int \frac{4 x + 1}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int 4 x + 1 d x

Apply u-substitution

: \int \frac{u}{4} d u

= \frac{1}{2} \cdot \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x + 1

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

Simplify

\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} (4 x + 1)^{\frac{3}{2}} : \frac{1}{12} (4 x + 1)^{\frac{3}{2}}

= \frac{1}{12} (4 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (4 x + 1)^{\frac{3}{2}} + C

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What is the integral of sqrt(x)sqrt(1+1/(4x)) ?
The integral of sqrt(x)sqrt(1+1/(4x)) is 1/12 (4x+1)^{3/2}+C