What is the integral of (3+12z)/(sqrt(2z^2+z+1)) ?

The integral of (3+12z)/(sqrt(2z^2+z+1)) is 6sqrt(2z^2+z+1)+C

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integral of (3+12z)/(sqrt(2z^2+z+1))

\int \frac{3 + 12 z}{2 z ^{2} + z + 1} d z

6 2 z^{2} + z + 1 + C

Solution steps

\int \frac{3 + 12 z}{2 z ^{2} + z + 1} d z

\frac{3 + 12 z}{2 z ^{2} + z + 1} = 3 \frac{1 + 4 z}{2 z ^{2} + z + 1}

= \int 3 \cdot \frac{1 + 4 z}{2 z ^{2} + z + 1} d z

Take the constant out:

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

= 3 \cdot \int \frac{1 + 4 z}{2 z ^{2} + z + 1} d z

Apply u-substitution

: \int \frac{1}{u} d u

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 2 z^{2} + z + 1

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

Simplify

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

= 6 2 z^{2} + z + 1

Add a constant to the solution

= 6 2 z^{2} + z + 1 + C

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What is the integral of (3+12z)/(sqrt(2z^2+z+1)) ?
The integral of (3+12z)/(sqrt(2z^2+z+1)) is 6sqrt(2z^2+z+1)+C