What is the integral of 1/((sqrt(a^2+r^2))^3) ?

The integral of 1/((sqrt(a^2+r^2))^3) is r/(a^2sqrt(a^2+r^2))+C

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integral of 1/((sqrt(a^2+r^2))^3)

\int \frac{1}{( a ^{2} + r ^{2} ) ^{3}} d r

\frac{r}{a ^{2} a ^{2} + r ^{2}} + C

Solution steps

\int \frac{1}{( a ^{2} + r ^{2} ) ^{3}} d r

Simplify

(a^{2} + r^{2})^{3} : (a^{2} + r^{2})^{2} a^{2} + r^{2}

= \int \frac{1}{( a ^{2} + r ^{2} ) ^{2} a ^{2} + r ^{2}} d r

Simplify

(a^{2} + r^{2})^{2} : a^{2} + r^{2} a^{2} + r^{2}

= \int \frac{1}{a ^{2} + r ^{2} a ^{2} + r ^{2} a ^{2} + r ^{2}} d r

Simplify

\frac{1}{a ^{2} + r ^{2} a ^{2} + r ^{2} a ^{2} + r ^{2}} : \frac{1}{( a ^{2} + r ^{2} ) ^{\frac{3}{2}}}

= \int \frac{1}{( a ^{2} + r ^{2} ) ^{\frac{3}{2}}} d r

Apply Trigonometric Substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

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

Use the common integral:

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

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

Substitute back

u = arctan (\frac{1}{a} r)

= \frac{1}{a ^{2}} sin (arctan (\frac{1}{a} r))

Simplify

\frac{1}{a ^{2}} sin (arctan (\frac{1}{a} r)) : \frac{r}{a ^{2} a ^{2} + r ^{2}}

= \frac{r}{a ^{2} a ^{2} + r ^{2}}

Add a constant to the solution

= \frac{r}{a ^{2} a ^{2} + r ^{2}} + C

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What is the integral of 1/((sqrt(a^2+r^2))^3) ?
The integral of 1/((sqrt(a^2+r^2))^3) is r/(a^2sqrt(a^2+r^2))+C