What is the integral of x/((4x^2-5)^6) ?

The integral of x/((4x^2-5)^6) is -1/(40(4x^2-5)^5)+C

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integral of x/((4x^2-5)^6)

\int \frac{x}{( 4 x ^{2} - 5 ) ^{6}} d x

- \frac{1}{40 ( 4 x ^{2} - 5 ) ^{5}} + C

Solution steps

\int \frac{x}{( 4 x ^{2} - 5 ) ^{6}} d x

Apply u-substitution

= \int \frac{1}{2 ( 4 u - 5 ) ^{6}} 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}{( 4 u - 5 ) ^{6}} d u

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{1}{4 v ^{6}} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{6}} d v

Apply the Power Rule

= \frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{5 v ^{5}})

Substitute back

= \frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{5 ( 4 x ^{2} - 5 ) ^{5}})

Simplify

\frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{5 ( 4 x ^{2} - 5 ) ^{5}}) : - \frac{1}{40 ( 4 x ^{2} - 5 ) ^{5}}

= - \frac{1}{40 ( 4 x ^{2} - 5 ) ^{5}}

Add a constant to the solution

= - \frac{1}{40 ( 4 x ^{2} - 5 ) ^{5}} + C

What is the integral of x/((4x^2-5)^6) ?
The integral of x/((4x^2-5)^6) is -1/(40(4x^2-5)^5)+C