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

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

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

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

6 x^{2} + 3 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 6 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

= 6 \cdot \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} + 3

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

Simplify

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

= 6 x^{2} + 3

Add a constant to the solution

= 6 x^{2} + 3 + C

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