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

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

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

\int \frac{x}{3 x ^{2} + 1} d x

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

(t + 4)^{\frac{2}{3}} (2 t^{2} - 3)^{3}

f (x) = ∣ x - 4 ∣, at x = 0

\frac{d}{d x} (0 x)

\frac{2 s + 2}{s ^{2} + 2 s + 3}

(1 + x)^{- 2}

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