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

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

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

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

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

Solution steps

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

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 = z^{2} + y^{2}

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

Simplify

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

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

Add a constant to the solution

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

(1, 7), (7, 3)

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\int \frac{1}{( x ln ( x ) )} d x

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