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

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

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

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

\frac{2}{3} 3 x^{2} + 2 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = 3 x^{2} + 2

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

Simplify

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

= \frac{2}{3} 3 x^{2} + 2

Add a constant to the solution

= \frac{2}{3} 3 x^{2} + 2 + C

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