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

The integral of r/(sqrt(r^2+z^2)) is sqrt(z^2+r^2)+C

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

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

z^{2} + r^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = z^{2} + r^{2}

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

Simplify

\frac{1}{2} \cdot 2 (z^{2} + r^{2})^{\frac{1}{2}} : z^{2} + r^{2}

= z^{2} + r^{2}

Add a constant to the solution

= z^{2} + r^{2} + C

\frac{d}{d x} (\frac{1}{1 + \frac{1}{x}})

x \to 3 lim (x^{2} + 5 x)

\frac{d y}{d x} = \frac{1}{11} y cos^{2} (y)

\frac{d y}{d x} = - 9 x - 9 y

f (x) = (4 x - 7)^{2}

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