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

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

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

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

\frac{1}{3} 2 + 3 x^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{6 u} d u

= \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

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

Apply the Power Rule

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

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

Substitute back

u = 2 + 3 x^{2}

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

Simplify

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

= \frac{1}{3} 2 + 3 x^{2}

Add a constant to the solution

= \frac{1}{3} 2 + 3 x^{2} + C

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