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

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

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

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

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

Solution steps

\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

= \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

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

x^{2}, - x + 5, 0

\frac{8 x ^{3} - 2 x + 9}{x}

\frac{\partial}{\partial u} (\frac{u}{u + v})

(x - 2) (3 x + 3)

f (x) = ln (1 + 2 x)

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