English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral de 1/(16x^4+8x^2+1)

Solución

\int \frac{1}{16 x ^{4} + 8 x ^{2} + 1} d x

\frac{1}{4} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x))) + C

Pasos de solución

\int \frac{1}{16 x ^{4} + 8 x ^{2} + 1} d x

Factorizar

\frac{1}{16 x ^{4} + 8 x ^{2} + 1}

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

Aplicar integración por sustitución

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

Re-escribir usando identidades trigonométricas

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

Sustitución hacia atrás

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

Simplificar

\frac{1}{2} \cdot \frac{1}{2} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x))) : \frac{1}{4} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x)))

= \frac{1}{4} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x)))

Agregar una constante a la solución

= \frac{1}{4} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x))) + C

\frac{\partial}{\partial x} (sin (k x))

\int (x + 2) x + 1 d x

y^{^{''}} = y + 4 t^{2}

x \to 5 lim (- 3 x)