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integral de 1/((4x^2-4x+5)^2)

Solución

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

\frac{1}{32} (arctan (x - \frac{1}{2}) + \frac{1}{2} sin (2 arctan (x - \frac{1}{2}))) + C

Pasos de solución

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

Completar el cuadrado

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

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

Aplicar integración por sustitución

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

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

Factorizar

\frac{1}{( 4 u ^{2} + 4 ) ^{2}}

= \int \frac{1}{16 ( u ^{2} + 1 ) ^{2}} d u

Sacar la constante:

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

= \frac{1}{16} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Aplicar integración por sustitución trigonométrica

: \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= \frac{1}{16} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Re-escribir usando identidades trigonométricas

= \frac{1}{16} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Sacar la constante:

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

= \frac{1}{16} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) 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

= \frac{1}{16} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{16} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Sustitución hacia atrás

= \frac{1}{16} \cdot \frac{1}{2} (arctan (x - \frac{1}{2}) + \frac{1}{2} sin (2 arctan (x - \frac{1}{2})))

Simplificar

\frac{1}{16} \cdot \frac{1}{2} (arctan (x - \frac{1}{2}) + \frac{1}{2} sin (2 arctan (x - \frac{1}{2}))) : \frac{1}{32} (arctan (x - \frac{1}{2}) + \frac{1}{2} sin (2 arctan (x - \frac{1}{2})))

= \frac{1}{32} (arctan (x - \frac{1}{2}) + \frac{1}{2} sin (2 arctan (x - \frac{1}{2})))

Agregar una constante a la solución

= \frac{1}{32} (arctan (x - \frac{1}{2}) + \frac{1}{2} sin (2 arctan (x - \frac{1}{2}))) + C

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