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

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

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integral of ((3x+1)^{1/2})/2

\int \frac{( 3 x + 1 ) ^{\frac{1}{2}}}{2} d x

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

Solution steps

\int \frac{( 3 x + 1 ) ^{\frac{1}{2}}}{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 (3 x + 1)^{\frac{1}{2}} d x

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{u ^{\frac{1}{2}}}{3} 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}{3} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

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

Substitute back

u = 3 x + 1

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

Simplify

\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{2}{3} (3 x + 1)^{\frac{3}{2}} : \frac{1}{9} (3 x + 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

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