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

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

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integral of 1/((y^2+z^2)^{3/2)}

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{z ^{2} sec ( u )} d u

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

Rewrite using trig identities

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

Use the common integral:

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 3 x^{2} + 2 x e^{2 x}

n = 9 \sum \infty e^{3 - 2 n}

\frac{1}{1 + x}

\int 5 x - 3 d x

x \to 0 lim (e^{π (x + 1)})

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