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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{y}{3 + y ^{2}} d y

Apply u-substitution

: \int \frac{1}{2 u} d u

= - \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 3 + y^{2}

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

Add a constant to the solution

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

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

\frac{\partial}{\partial x} (t^{6} e^{- x})

y^{^{''}} + 4 y = 4 x

\int x^{2} + 5 x + 1 d x

\frac{d}{d x} f (x) = (x - 1 - x^{2})^{2}

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