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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

(- 8.8) (- 4.3)

\int e^{5 x} sin (x) d x

x y^{^{'}} = y + 2 x^{2} sin (x)

\frac{d}{d x} (g (x))

\int_{1}^{2} x^{3} (x^{2} - 4) d x

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