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

The integral of (4x-2)/((4x^2-4x+2)^4) is -1/(48(2x^2-2x+1)^3)+C

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

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

- \frac{1}{48 ( 2 x ^{2} - 2 x + 1 ) ^{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{64 u}{( u ^{2} + 4 ) ^{4}} d u

= \int \frac{64 u}{( u ^{2} + 4 ) ^{4}} d u

Take the constant out:

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

= 64 \cdot \int \frac{u}{( u ^{2} + 4 ) ^{4}} d u

Apply u-substitution

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

= 64 \cdot \int \frac{1}{2 v ^{4}} d v

Take the constant out:

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

= 64 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

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

= 64 \cdot \frac{1}{2} (- \frac{1}{3 v ^{3}})

Substitute back

= 64 \cdot \frac{1}{2} - \frac{1}{3 ( ( 4 x - 2 ) ^{2} + 4 ) ^{3}}

Simplify

64 \cdot \frac{1}{2} - \frac{1}{3 ( ( 4 x - 2 ) ^{2} + 4 ) ^{3}} : - \frac{1}{48 ( 2 x ^{2} - 2 x + 1 ) ^{3}}

= - \frac{1}{48 ( 2 x ^{2} - 2 x + 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{48 ( 2 x ^{2} - 2 x + 1 ) ^{3}} + C

\frac{1 - e ^{- 2 s}}{6 s ^{2}}

x \to 0 lim (- cos (x) + 1)

\int \frac{5 x ^{2}}{x + 1} d x

y = x (1 - x)^{6}

f (s) = s - 3 (s^{2} + 4 s + 13)

What is the integral of (4x-2)/((4x^2-4x+2)^4) ?
The integral of (4x-2)/((4x^2-4x+2)^4) is -1/(48(2x^2-2x+1)^3)+C