What is the integral of 1/((sqrt(x^2+y^2))^3) ?

The integral of 1/((sqrt(x^2+y^2))^3) is y/(x^2sqrt(x^2+y^2))+C

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integral of 1/((sqrt(x^2+y^2))^3)

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

\frac{y}{x ^{2} x ^{2} + y ^{2}} + C

Solution steps

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

Simplify

(x^{2} + y^{2})^{3} : (x^{2} + y^{2})^{2} x^{2} + y^{2}

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

Simplify

(x^{2} + y^{2})^{2} : x^{2} + y^{2} x^{2} + y^{2}

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

Simplify

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

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

Apply Trigonometric Substitution

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

Rewrite using trig identities

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

Use the common integral:

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

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

Substitute back

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

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

Simplify

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

= \frac{y}{x ^{2} x ^{2} + y ^{2}}

Add a constant to the solution

= \frac{y}{x ^{2} x ^{2} + y ^{2}} + C

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