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

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

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

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

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

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

Simplify

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

= x^{2} + y^{2}

Add a constant to the solution

= x^{2} + y^{2} + C

\frac{\partial}{\partial x} (x y^{2} - y x^{3})

\frac{d}{d x} (1 + 40 x^{3} - 3 x^{5})

x \to 2 lim (1680 x^{3} - 120 x)

\int \frac{2 x - 5}{x ^{2} + 2 x + 2} d x

\int \frac{1}{8 x} d x

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