What is the integral of ((8x-4))/(4x^2-4x+2) ?

The integral of ((8x-4))/(4x^2-4x+2) is ln|4x^2-4x+2|+C

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integral of ((8x-4))/(4x^2-4x+2)

\int \frac{( 8 x - 4 )}{4 x ^{2} - 4 x + 2} d x

ln 4 x^{2} - 4 x + 2 + C

Solution steps

\int \frac{8 x - 4}{4 x ^{2} - 4 x + 2} d x

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 4 x^{2} - 4 x + 2

= ln 4 x^{2} - 4 x + 2

Add a constant to the solution

= ln 4 x^{2} - 4 x + 2 + C

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

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\int (4 x + 4) d x

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\int 0.5 - cos (x) d x

What is the integral of ((8x-4))/(4x^2-4x+2) ?
The integral of ((8x-4))/(4x^2-4x+2) is ln|4x^2-4x+2|+C