What is the integral of (-2x-1)/((x^2+x+1)^6) ?

The integral of (-2x-1)/((x^2+x+1)^6) is 1/(5(x^2+x+1)^5)+C

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integral of (-2x-1)/((x^2+x+1)^6)

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

\frac{1}{5 ( x ^{2} + x + 1 ) ^{5}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{2048 u}{( u ^{2} + 3 ) ^{6}} d u

= \int - \frac{2048 u}{( u ^{2} + 3 ) ^{6}} d u

Take the constant out:

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

= - 2048 \cdot \int \frac{u}{( u ^{2} + 3 ) ^{6}} d u

Apply u-substitution

: \int \frac{1}{2 v ^{6}} d v

= - 2048 \cdot \int \frac{1}{2 v ^{6}} d v

Take the constant out:

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

= - 2048 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{6}} d v

Apply the Power Rule

: - \frac{1}{5 v ^{5}}

= - 2048 \cdot \frac{1}{2} (- \frac{1}{5 v ^{5}})

Substitute back

= - 2048 \cdot \frac{1}{2} - \frac{1}{5 ( ( - 2 x - 1 ) ^{2} + 3 ) ^{5}}

Simplify

- 2048 \cdot \frac{1}{2} - \frac{1}{5 ( ( - 2 x - 1 ) ^{2} + 3 ) ^{5}} : \frac{1}{5 ( x ^{2} + x + 1 ) ^{5}}

= \frac{1}{5 ( x ^{2} + x + 1 ) ^{5}}

Add a constant to the solution

= \frac{1}{5 ( x ^{2} + x + 1 ) ^{5}} + C

\frac{d y}{d x} = \frac{y ^{2}}{x ^{2}}

\frac{\partial}{\partial x} (12 cos (6 x) cos (4 y))

\int_{0}^{3} \frac{1}{x + 1} d x

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

What is the integral of (-2x-1)/((x^2+x+1)^6) ?
The integral of (-2x-1)/((x^2+x+1)^6) is 1/(5(x^2+x+1)^5)+C