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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{y}{1 + x ^{2}}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{3 x}{4})

\frac{1}{4 x}

h \to 0 lim (- 2 - h)

y^{^{'}} = k (80 - y) (50 - y)

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