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

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

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integral of sqrt(x+1/4)

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

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

Solution steps

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

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

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

Substitute back

u = x + \frac{1}{4}

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

Simplify

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