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

The integral of x/((x^2-9)^5) is -1/(8(x^2-9)^4)+C

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

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

- \frac{1}{8 ( x ^{2} - 9 ) ^{4}} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{4 v ^{4}}

= \frac{1}{2} (- \frac{1}{4 v ^{4}})

Substitute back

= \frac{1}{2} (- \frac{1}{4 ( x ^{2} - 9 ) ^{4}})

Simplify

\frac{1}{2} (- \frac{1}{4 ( x ^{2} - 9 ) ^{4}}) : - \frac{1}{8 ( x ^{2} - 9 ) ^{4}}

= - \frac{1}{8 ( x ^{2} - 9 ) ^{4}}

Add a constant to the solution

= - \frac{1}{8 ( x ^{2} - 9 ) ^{4}} + C

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What is the integral of x/((x^2-9)^5) ?
The integral of x/((x^2-9)^5) is -1/(8(x^2-9)^4)+C