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

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

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

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

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

Solution steps

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 1 + 3 y^{2}

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

Add a constant to the solution

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

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