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integral de (-x+2)/(x^2-x+1)

Solución

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

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

Pasos de solución

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

Completar el cuadrado

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

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

Aplicar integración por sustitución

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

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

Sacar la constante:

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

= 2 \cdot \int \frac{- 2 u + 3}{4 u ^{2} + 3} d u

Aplicar integración por sustitución

= 2 \cdot \int \frac{- v + 3}{2 ( v ^{2} + 1 )} d v

Sacar la constante:

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

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

Expandir

\frac{- v + 3}{v ^{2} + 1} : - \frac{v}{v ^{2} + 1} + \frac{3}{v ^{2} + 1}

Aplicar la regla de la suma:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{2} (- \int \frac{v}{v ^{2} + 1} d v + \int \frac{3}{v ^{2} + 1} d v)

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= 2 \cdot \frac{1}{2} (- \frac{1}{2} ln v^{2} + 1 + 3 arctan (v))

Sustitución hacia atrás

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

Simplificar

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

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

Agregar una constante a la solución

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

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\frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}}

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\frac{\partial}{\partial x} (r + 2 r^{4} x + 4 x)

(4 + y^{2}) d x + (9 + x^{2}) d y = 0