What is the integral of (3x-2)/((3x^2-4x+2)^3) ?

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

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integral of (3x-2)/((3x^2-4x+2)^3)

\int \frac{3 x - 2}{( 3 x ^{2} - 4 x + 2 ) ^{3}} d x

- \frac{1}{4 ( 3 x ^{2} - 4 x + 2 ) ^{2}} + C

Solution steps

\int \frac{3 x - 2}{( 3 x ^{2} - 4 x + 2 ) ^{3}} d x

Apply u-substitution

: \int \frac{9 u}{( u ^{2} + 2 ) ^{3}} d u

= \int \frac{9 u}{( u ^{2} + 2 ) ^{3}} d u

Take the constant out:

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

= 9 \cdot \int \frac{u}{( u ^{2} + 2 ) ^{3}} d u

Apply u-substitution

: \int \frac{1}{2 v ^{3}} d v

= 9 \cdot \int \frac{1}{2 v ^{3}} d v

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{3}} d v

Apply the Power Rule

: - \frac{1}{2 v ^{2}}

= 9 \cdot \frac{1}{2} (- \frac{1}{2 v ^{2}})

Substitute back

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

Simplify

9 \cdot \frac{1}{2} - \frac{1}{2 ( ( 3 x - 2 ) ^{2} + 2 ) ^{2}} : - \frac{1}{4 ( 3 x ^{2} - 4 x + 2 ) ^{2}}

= - \frac{1}{4 ( 3 x ^{2} - 4 x + 2 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( 3 x ^{2} - 4 x + 2 ) ^{2}} + C

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