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

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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{x ( 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}{x} \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}{x} arctan (u)

Substitute back

u = \frac{y}{x}

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

Add a constant to the solution

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

x^{0.5}

\frac{d}{d x} (x sin (π x))

\frac{d}{d x} (9 e^{x} (cos (x) - sin (x)))

x \to 0 lim ((1 + 6 x)^{\frac{7}{x}})

2 x - tan (x)

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