What is the integral of (4x-2)/((x^2-x+1)^7) ?

The integral of (4x-2)/((x^2-x+1)^7) is -1/(3(x^2-x+1)^6)+C

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integral of (4x-2)/((x^2-x+1)^7)

\int \frac{4 x - 2}{( x ^{2} - x + 1 ) ^{7}} d x

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

Solution steps

\int \frac{4 x - 2}{( x ^{2} - x + 1 ) ^{7}} d x

Apply u-substitution

: \int \frac{2}{u ^{7}} d u

= \int \frac{2}{u ^{7}} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u ^{7}} d u

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= 2 (- \frac{1}{6 u ^{6}})

Substitute back

u = x^{2} - x + 1

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

Simplify

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

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

Add a constant to the solution

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

f (x) = ln (cos (2 x))

\int \frac{8}{16 - x ^{2}} d x

y = 7 - x, x = \frac{( y - 4 ) ^{2}}{4}

\int (2 x + 1) e^{x} d x

\frac{\partial}{\partial x} (6 x + 3 y)

What is the integral of (4x-2)/((x^2-x+1)^7) ?
The integral of (4x-2)/((x^2-x+1)^7) is -1/(3(x^2-x+1)^6)+C