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

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

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integral of z/((s^2+z^2)^{3/2)}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{3}{2}}} d u

= \int \frac{1}{2 u ^{\frac{3}{2}}} 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 ^{\frac{3}{2}}} d u

Apply the Power Rule

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

= \frac{1}{2} (- \frac{2}{u ^{\frac{1}{2}}})

Substitute back

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

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

Simplify

\frac{1}{2} (- \frac{2}{( s ^{2} + z ^{2} ) ^{\frac{1}{2}}}) : - \frac{1}{( s ^{2} + z ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

m (x) = \frac{- x ^{2} + 6 x - 8}{- x ^{2} + 6 x - 2}

y = - 5 - 6 x^{2}, (- 4, - 101)

p (t) = 5 e^{0.0278 t}

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

\frac{d}{d x} (5 x e^{5 x})

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