What is the integral of (4+6x)/(sqrt(6+4x+3x^2)) ?

The integral of (4+6x)/(sqrt(6+4x+3x^2)) is 2sqrt(6+4x+3x^2)+C

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integral of (4+6x)/(sqrt(6+4x+3x^2))

\int \frac{4 + 6 x}{6 + 4 x + 3 x ^{2}} d x

2 6 + 4 x + 3 x^{2} + C

Solution steps

\int \frac{4 + 6 x}{6 + 4 x + 3 x ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = 6 + 4 x + 3 x^{2}

= 2 (6 + 4 x + 3 x^{2})^{\frac{1}{2}}

Apply radical rule:

= 2 6 + 4 x + 3 x^{2}

Add a constant to the solution

= 2 6 + 4 x + 3 x^{2} + C

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What is the integral of (4+6x)/(sqrt(6+4x+3x^2)) ?
The integral of (4+6x)/(sqrt(6+4x+3x^2)) is 2sqrt(6+4x+3x^2)+C