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

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

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

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

\frac{3}{2} ln z^{2} + y^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

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

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

Simplify

3 \cdot \frac{1}{2} ln z^{2} + y^{2} : \frac{3}{2} ln z^{2} + y^{2}

= \frac{3}{2} ln z^{2} + y^{2}

Add a constant to the solution

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

(x - 1)^{12}

\int x - x^{2} d x

81 y^{^{''}} - 18 y^{^{'}} + y = 0

\int x^{- 2} sin (\frac{1}{x}) d x

\frac{d P}{d t} = P t, P (1) = 2

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