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

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

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

integral

\frac{y}{x ^{2} + y ^{2}}

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

Solution steps

\int \frac{y}{x ^{2} + 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 = x^{2} + y^{2}

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

Add a constant to the solution

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

2 + x

\frac{1}{( x ^{\frac{1}{2}} + 1 )}

(x^{2} - x)^{4}

\frac{2}{( 2 x + 3 )}

\frac{x}{( x + 2 ) ^{2}}

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