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

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

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

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

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

Solution steps

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

Take the constant out:

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

= c \cdot \int \frac{y}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}} d y

Apply u-substitution

= c \cdot \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

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

Apply the Power Rule

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

Substitute back

u = x^{2} + y^{2}

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

Simplify

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

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

Add a constant to the solution

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

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