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

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

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

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

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

Solution steps

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

Simplify

(y^{2} + x^{2})^{\frac{- 3}{2}} : \frac{1}{( y ^{2} + x ^{2} ) ^{\frac{3}{2}}}

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

Apply Trigonometric Substitution

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

Rewrite using trig identities

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

Use the common integral:

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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