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

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

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

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

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

Solution steps

\int \frac{x}{( x ^{2} + z ^{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 = z^{2} + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x e^{x y} + y^{2} sin (x y))

\frac{\partial}{\partial x} (ln (x^{7} + y^{4}))

\int cot (\frac{θ}{3}) d θ

\frac{2}{s ^{3}}

cos^{x} (x)

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