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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= \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{1}{2}}}

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

Add a constant to the solution

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

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

y = 2 x^{2} + 1

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3} + \frac{1}{4 x + 4})

3 x^{2} e^{x}

\int 2 sin (x) + x d x

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