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

The integral of sqrt(x^2+y^2+z^2) is

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integral of sqrt(x^2+y^2+z^2)

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int sec^{3} (u) (x^{2} + y^{2}) d u

= \int sec^{3} (u) (x^{2} + y^{2}) d u

Take the constant out:

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

= (x^{2} + y^{2}) \cdot \int sec^{3} (u) d u

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= (x^{2} + y^{2}) (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

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

Simplify

(x^{2} + y^{2}) (\frac{1}{2} sec (arctan (\frac{1}{x ^{2} + y ^{2}} z)) tan (arctan (\frac{1}{x ^{2} + y ^{2}} z)) + \frac{1}{2} ln tan (arctan (\frac{1}{x ^{2} + y ^{2}} z)) + sec (arctan (\frac{1}{x ^{2} + y ^{2}} z))) : (x^{2} + y^{2}) (\frac{z x ^{2} + y ^{2} + z ^{2}}{2 ( x ^{2} + y ^{2} )} + \frac{1}{2} ln \frac{z}{x ^{2} + y ^{2}} + \frac{x ^{2} + y ^{2} + z ^{2}}{x ^{2} + y ^{2}})

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

Add a constant to the solution

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

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