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

The integral of (6z+2)^{1/2} is (2*2^{1/2})/9 (3z+1)^{3/2}+C

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

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

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

Solution steps

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

(6 z + 2)^{\frac{1}{2}} = 2^{\frac{1}{2}} (3 z + 1)^{\frac{1}{2}}

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

Take the constant out:

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

= 2^{\frac{1}{2}} \cdot \int (3 z + 1)^{\frac{1}{2}} d z

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{3} d u

= 2^{\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

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

Apply the Power Rule

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

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

Substitute back

u = 3 z + 1

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (6z+2)^{1/2} ?
The integral of (6z+2)^{1/2} is (2*2^{1/2})/9 (3z+1)^{3/2}+C