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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply u-substitution

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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