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

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

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

integral

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

\frac{1}{y} arctan (\frac{x}{y}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{y ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{y} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{y} arctan (u)

Substitute back

u = \frac{x}{y}

= \frac{1}{y} arctan (\frac{x}{y})

Add a constant to the solution

= \frac{1}{y} arctan (\frac{x}{y}) + C

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