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

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

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

integral

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

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

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Substitute back

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

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

Add a constant to the solution

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

\frac{( x ^{2} - 2 )}{( x + 4 )}

(\frac{1}{x}) sin (\frac{1}{x})

\frac{- 2}{y}

8 x - x^{2}

(\frac{x ^{2} - 1}{x})^{3}

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