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

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

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integral of y/(y^2+z^2)

\int \frac{y}{y ^{2} + z ^{2}} d y

\frac{1}{2} ln z^{2} + y^{2} + C

Solution steps

\int \frac{y}{y ^{2} + z ^{2}} d y

Apply u-substitution

= \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 = z^{2} + y^{2}

= \frac{1}{2} ln z^{2} + y^{2}

Add a constant to the solution

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

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