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

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

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integral of x/(sqrt(3x^2+4))

\int \frac{x}{3 x ^{2} + 4} d x

\frac{1}{3} 3 x^{2} + 4 + C

Solution steps

\int \frac{x}{3 x ^{2} + 4} 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} + 4

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

Simplify

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

= \frac{1}{3} 3 x^{2} + 4

Add a constant to the solution

= \frac{1}{3} 3 x^{2} + 4 + C

f (x) = \frac{x}{2 x - 4}, at x = 5

n = 1 \sum \infty \frac{2}{1 + n ^{2}}

\frac{d}{d x} (- 2 x + 4)

\frac{1}{2 x + 3}

\frac{d}{d x} (2 + sec (x))

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