What is the integral of 1/(sqrt((x^2+9)^3)) ?

The integral of 1/(sqrt((x^2+9)^3)) is x/(9sqrt(9+x^2))+C

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integral of 1/(sqrt((x^2+9)^3))

\int \frac{1}{( x ^{2} + 9 ) ^{3}} d x

\frac{x}{9 9 + x ^{2}} + C

Solution steps

\int \frac{1}{( x ^{2} + 9 ) ^{3}} d x

(x^{2} + 9)^{3} = (x^{2} + 9)^{\frac{3}{2}},

assuming

(x^{2} + 9) \geq 0

= \int \frac{1}{( x ^{2} + 9 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{9 sec ( u )} d u

= \int \frac{1}{9 sec ( u )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{9} sin (u)

Substitute back

u = arctan (\frac{1}{3} x)

= \frac{1}{9} sin (arctan (\frac{1}{3} x))

Simplify

\frac{1}{9} sin (arctan (\frac{1}{3} x)) : \frac{x}{9 9 + x ^{2}}

= \frac{x}{9 9 + x ^{2}}

Add a constant to the solution

= \frac{x}{9 9 + x ^{2}} + C

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What is the integral of 1/(sqrt((x^2+9)^3)) ?
The integral of 1/(sqrt((x^2+9)^3)) is x/(9sqrt(9+x^2))+C