What is the integral of 1/((ρ^2+z^2)^{3/2)} ?

The integral of 1/((ρ^2+z^2)^{3/2)} is z/(ρ^2(ρ^2+z^2)^{1/2)}+C

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integral of 1/((ρ^2+z^2)^{3/2)}

\int \frac{1}{( ρ ^{2} + z ^{2} ) ^{\frac{3}{2}}} d z

\frac{z}{ρ ^{2} ( ρ ^{2} + z ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( ρ ^{2} + z ^{2} ) ^{\frac{3}{2}}} d z

Apply Trigonometric Substitution

: \int \frac{1}{ρ ^{2} sec ( u )} d u

= \int \frac{1}{ρ ^{2} sec ( u )} d u

Take the constant out:

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

= \frac{1}{ρ ^{2}} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{ρ ^{2}} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{ρ ^{2}} sin (u)

Substitute back

u = arctan (\frac{1}{ρ} z)

= \frac{1}{ρ ^{2}} sin (arctan (\frac{1}{ρ} z))

Simplify

\frac{1}{ρ ^{2}} sin (arctan (\frac{1}{ρ} z)) : \frac{z}{ρ ^{2} ( ρ ^{2} + z ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

= \frac{z}{ρ ^{2} ( ρ ^{2} + z ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{1}{( 1 + x ^{2} ) x} d x

sin (y) \frac{d y}{d x} - cos (y) = 0

\frac{d}{d x} (\frac{x - 3}{2})

\int 6 x^{5} - 4 x^{4} - 9 x^{2} d x

\int (2 x^{3} - 9 x^{2} + 5 x) e^{- x} d x

What is the integral of 1/((ρ^2+z^2)^{3/2)} ?
The integral of 1/((ρ^2+z^2)^{3/2)} is z/(ρ^2(ρ^2+z^2)^{1/2)}+C