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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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

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

Substitute back

u = 3 s^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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