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

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

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

integral

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

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

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{y ^{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)

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

Substitute back

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

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

Add a constant to the solution

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

in t e g r a l \frac{1}{tan ^{2} ( x )}

in t e g r a l \frac{x ^{3}}{( 1 + x ^{2} ) ^{2}}

in t e g r a l sin^{2} (x) + cos^{2} (x)

in t e g r a l \frac{ln ( x )}{( x + 1 ) ^{2}}

in t e g r a l \frac{sin ^{2} ( x )}{2}

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