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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 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

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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