What is the integral of 1/((2x-9)^4) ?

The integral of 1/((2x-9)^4) is -1/(6(2x-9)^3)+C

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integral of 1/((2x-9)^4)

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

- \frac{1}{6 ( 2 x - 9 ) ^{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{4}} d u

= \int \frac{1}{2 u ^{4}} 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 ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{2} (- \frac{1}{3 u ^{3}})

Substitute back

u = 2 x - 9

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

Simplify

\frac{1}{2} (- \frac{1}{3 ( 2 x - 9 ) ^{3}}) : - \frac{1}{6 ( 2 x - 9 ) ^{3}}

= - \frac{1}{6 ( 2 x - 9 ) ^{3}}

Add a constant to the solution

= - \frac{1}{6 ( 2 x - 9 ) ^{3}} + C

5 y^{^{''}} + 19 y^{^{'}} - 4 y = 0

\int sin^{3} (y) cos (y) d y

f (x) = (1 + x)^{5}

g (x) = (1 - 3 x)

\frac{d}{d x} (8 - 3 x)

What is the integral of 1/((2x-9)^4) ?
The integral of 1/((2x-9)^4) is -1/(6(2x-9)^3)+C