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

Solución

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

2 (\frac{1}{5} ln ∣ 2 x - 1 ∣ + \frac{1}{10} (- ln 4 x^{2} + 4 - arctan (x))) + C

Pasos de solución

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

Aplicar integración por sustitución

: \int \frac{2}{u ( u ^{2} + 2 u + 5 )} d u

= \int \frac{2}{u ( u ^{2} + 2 u + 5 )} d u

Sacar la constante:

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

= 2 \cdot \int \frac{1}{u ( u ^{2} + 2 u + 5 )} d u

Completar el cuadrado

u^{2} + 2 u + 5 : (u + 1)^{2} + 4

= 2 \cdot \int \frac{1}{u ( ( u + 1 ) ^{2} + 4 )} d u

Aplicar integración por sustitución

: \int \frac{1}{( v - 1 ) ( v ^{2} + 4 )} d v

= 2 \cdot \int \frac{1}{( v - 1 ) ( v ^{2} + 4 )} d v

Tomar la fracción parcial de

\frac{1}{( v - 1 ) ( v ^{2} + 4 )} : \frac{1}{5 ( v - 1 )} + \frac{- v - 1}{5 ( v ^{2} + 4 )}

= 2 \cdot \int \frac{1}{5 ( v - 1 )} + \frac{- v - 1}{5 ( v ^{2} + 4 )} d v

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 (\int \frac{1}{5 ( v - 1 )} d v + \int \frac{- v - 1}{5 ( v ^{2} + 4 )} d v)

\int \frac{1}{5 ( v - 1 )} d v = \frac{1}{5} ln ∣ v - 1 ∣

\int \frac{- v - 1}{5 ( v ^{2} + 4 )} d v = \frac{1}{5} (- \frac{1}{2} ln v^{2} + 4 - \frac{1}{2} arctan (\frac{v}{2}))

= 2 (\frac{1}{5} ln ∣ v - 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln v^{2} + 4 - \frac{1}{2} arctan (\frac{v}{2})))

Sustitución hacia atrás

= 2 (\frac{1}{5} ln ∣ 2 x - 1 + 1 - 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln (2 x - 1 + 1)^{2} + 4 - \frac{1}{2} arctan (\frac{2 x - 1 + 1}{2})))

Simplificar

2 (\frac{1}{5} ln ∣ 2 x - 1 + 1 - 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln (2 x - 1 + 1)^{2} + 4 - \frac{1}{2} arctan (\frac{2 x - 1 + 1}{2}))) : 2 (\frac{1}{5} ln ∣ 2 x - 1 ∣ + \frac{1}{10} (- ln 4 x^{2} + 4 - arctan (x)))

= 2 (\frac{1}{5} ln ∣ 2 x - 1 ∣ + \frac{1}{10} (- ln 4 x^{2} + 4 - arctan (x)))

Agregar una constante a la solución

= 2 (\frac{1}{5} ln ∣ 2 x - 1 ∣ + \frac{1}{10} (- ln 4 x^{2} + 4 - arctan (x))) + C

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