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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= r^{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) ∣

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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\frac{\partial}{\partial y} (2 x + 3 y - 7)

\int 0.8 x + 3 d x

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