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

The integral of (3x)/(sqrt(6x^2-4)) is 1/2 sqrt(6x^2-4)+C

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

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

\frac{1}{2} 6 x^{2} - 4 + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{x}{6 x ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{12 u} d u

= 3 \cdot \int \frac{1}{12 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{12} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{12} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 6 x^{2} - 4

= 3 \cdot \frac{1}{12} \cdot 2 (6 x^{2} - 4)^{\frac{1}{2}}

Simplify

3 \cdot \frac{1}{12} \cdot 2 (6 x^{2} - 4)^{\frac{1}{2}} : \frac{1}{2} 6 x^{2} - 4

= \frac{1}{2} 6 x^{2} - 4

Add a constant to the solution

= \frac{1}{2} 6 x^{2} - 4 + C

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What is the integral of (3x)/(sqrt(6x^2-4)) ?
The integral of (3x)/(sqrt(6x^2-4)) is 1/2 sqrt(6x^2-4)+C