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

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

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

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

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d y = y

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

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

Add a constant to the solution

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

f (x) = 8 x^{2} - x^{3}, at x = 2

\frac{\partial}{\partial x} (\frac{y}{2 x})

\frac{\partial}{\partial x} (- 4 x y + x^{4} + y^{4})

y^{^{''}} + 4 y = 0

\int x sin (1 - 2 x) d x

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