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

The integral of (1+x^2)/(1+y^2) is 1/(1+y^2)(x+(x^3)/3)+C

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

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

\frac{1}{1 + y ^{2}} (x + \frac{x ^{3}}{3}) + C

Solution steps

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

Take the constant out:

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

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

Apply the Sum Rule:

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

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

\int 1 d x = x

\int x^{2} d x = \frac{x ^{3}}{3}

= \frac{1}{1 + y ^{2}} (x + \frac{x ^{3}}{3})

Add a constant to the solution

= \frac{1}{1 + y ^{2}} (x + \frac{x ^{3}}{3}) + C

(x^{2} + 3 y^{2}) d x - 2 x y d y = 0

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

\int_{0}^{2 π} cos (x) sin (x) d x

x \to - 6 lim (11 x + 3)

f (x) = 12 - x^{2}, (3, 3)

What is the integral of (1+x^2)/(1+y^2) ?
The integral of (1+x^2)/(1+y^2) is 1/(1+y^2)(x+(x^3)/3)+C