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

The integral of y/(3y^2+1) is 1/6 ln|3y^2+1|+C

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integral of y/(3y^2+1)

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

\frac{1}{6} ln 3 y^{2} + 1 + C

Solution steps

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 y^{2} + 1

= \frac{1}{6} ln 3 y^{2} + 1

Add a constant to the solution

= \frac{1}{6} ln 3 y^{2} + 1 + C

What is the integral of y/(3y^2+1) ?
The integral of y/(3y^2+1) is 1/6 ln|3y^2+1|+C