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

The integral of sqrt(x^2+z^2) is 1/2 zsqrt(z^2+x^2)+(x^2)/2 ln((|z+sqrt(x^2+z^2)|)/(|x|))+C

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

Substitute back

u = arctan (\frac{1}{x} z)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sqrt(x^2+z^2) ?
The integral of sqrt(x^2+z^2) is 1/2 zsqrt(z^2+x^2)+(x^2)/2 ln((|z+sqrt(x^2+z^2)|)/(|x|))+C