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

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

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

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

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

Solution steps

\int \frac{x}{( 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

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

Apply Trigonometric Substitution

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

= x \cdot \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

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

Rewrite using trig identities

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

Use the common integral:

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{5} \frac{1}{( 5 - x ) ^{\frac{5}{2}}} d x

\frac{\partial}{\partial x} (y^{3} - 3 x^{2} y)

f (x) = (4 x + 5)^{3} (x^{2} - 6 x + 5)^{4}

\int \frac{36}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

(- 5) (- 5.1)

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