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

Apply exponent rule:

(a^{n})^{m} = a^{n \cdot m}

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

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

Multiply by the conjugate of

x^{2} + 9 : \frac{x ^{4} - 81}{x ^{2} - 9}

= \int \frac{1}{( \frac{x ^{4} - 81}{x ^{2} - 9} ) ^{\frac{3}{2}}} d x

Apply Integral Substitution

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

Apply Trigonometric Substitution

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

Use the common integral:

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

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

Substitute back

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

Simplify

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