What is the integral of (16x)/((sqrt(x^2-9))^5) ?

The integral of (16x)/((sqrt(x^2-9))^5) is -(16)/(3(x^2-9)^{3/2)}+C

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integral of (16x)/((sqrt(x^2-9))^5)

\int \frac{16 x}{( x ^{2} - 9 ) ^{5}} d x

- \frac{16}{3 ( x ^{2} - 9 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{16 x}{( x ^{2} - 9 ) ^{5}} d x

Take the constant out:

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

= 16 \cdot \int \frac{x}{( x ^{2} - 9 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{2 ( u - 9 ) ^{\frac{5}{2}}} d u

= 16 \cdot \int \frac{1}{2 ( u - 9 ) ^{\frac{5}{2}}} d u

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 9 ) ^{\frac{5}{2}}} d u

Apply u-substitution

: \int \frac{1}{v ^{\frac{5}{2}}} d v

= 16 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{5}{2}}} d v

Apply the Power Rule

: - \frac{2}{3 v ^{\frac{3}{2}}}

= 16 \cdot \frac{1}{2} (- \frac{2}{3 v ^{\frac{3}{2}}})

Substitute back

= 16 \cdot \frac{1}{2} (- \frac{2}{3 ( x ^{2} - 9 ) ^{\frac{3}{2}}})

Simplify

16 \cdot \frac{1}{2} (- \frac{2}{3 ( x ^{2} - 9 ) ^{\frac{3}{2}}}) : - \frac{16}{3 ( x ^{2} - 9 ) ^{\frac{3}{2}}}

= - \frac{16}{3 ( x ^{2} - 9 ) ^{\frac{3}{2}}}

Add a constant to the solution

= - \frac{16}{3 ( x ^{2} - 9 ) ^{\frac{3}{2}}} + C

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What is the integral of (16x)/((sqrt(x^2-9))^5) ?
The integral of (16x)/((sqrt(x^2-9))^5) is -(16)/(3(x^2-9)^{3/2)}+C