What is the integral of ((5x-6))/((2x+3)^9) ?

The integral of ((5x-6))/((2x+3)^9) is (-80x+69)/(224(2x+3)^8)+C

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integral of ((5x-6))/((2x+3)^9)

\int \frac{( 5 x - 6 )}{( 2 x + 3 ) ^{9}} d x

\frac{- 80 x + 69}{224 ( 2 x + 3 ) ^{8}} + C

Solution steps

\int \frac{5 x - 6}{( 2 x + 3 ) ^{9}} d x

Apply u-substitution

: \int \frac{5 u - 27}{4 u ^{9}} d u

= \int \frac{5 u - 27}{4 u ^{9}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{5 u - 27}{u ^{9}} d u

Expand

\frac{5 u - 27}{u ^{9}} : \frac{5}{u ^{8}} - \frac{27}{u ^{9}}

= \frac{1}{4} \cdot \int \frac{5}{u ^{8}} - \frac{27}{u ^{9}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int \frac{5}{u ^{8}} d u - \int \frac{27}{u ^{9}} d u)

\int \frac{5}{u ^{8}} d u = - \frac{5}{7 u ^{7}}

\int \frac{27}{u ^{9}} d u = - \frac{27}{8 u ^{8}}

= \frac{1}{4} (- \frac{5}{7 u ^{7}} - (- \frac{27}{8 u ^{8}}))

Substitute back

u = 2 x + 3

= \frac{1}{4} (- \frac{5}{7 ( 2 x + 3 ) ^{7}} - (- \frac{27}{8 ( 2 x + 3 ) ^{8}}))

Simplify

\frac{1}{4} (- \frac{5}{7 ( 2 x + 3 ) ^{7}} - (- \frac{27}{8 ( 2 x + 3 ) ^{8}})) : \frac{- 80 x + 69}{224 ( 2 x + 3 ) ^{8}}

= \frac{- 80 x + 69}{224 ( 2 x + 3 ) ^{8}}

Add a constant to the solution

= \frac{- 80 x + 69}{224 ( 2 x + 3 ) ^{8}} + C

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What is the integral of ((5x-6))/((2x+3)^9) ?
The integral of ((5x-6))/((2x+3)^9) is (-80x+69)/(224(2x+3)^8)+C