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integral de 1/(sqrt(12x+0.02x^2))

Solución

\int \frac{1}{12 x + 0.02 x ^{2}} d x

7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6) + C

Pasos de solución

\int \frac{1}{12 x + 0.02 x ^{2}} d x

Factorizar

\frac{1}{12 x + 0.02 x ^{2}}

= \int \frac{1}{x ( 0.02 x + 12 )} d x

Aplicar integración por sustitución

: \int \frac{1}{2 \cdot 0.1 u ( u - 12 )} d u

= \int \frac{1}{2 \cdot 0.1 u ( u - 12 )} d u

Sacar la constante:

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

= \frac{1}{2 \cdot 0.1} \cdot \int \frac{1}{u ( u - 12 )} d u

Simplificar

= 7.07106 \dots \cdot \int \frac{1}{u ( u - 12 )} d u

Completar el cuadrado

u (u - 12) : (u - 6)^{2} - 36

= 7.07106 \dots \cdot \int \frac{1}{( u - 6 ) ^{2} - 36} d u

Aplicar integración por sustitución

: \int \frac{1}{v ^{2} - 36} d v

= 7.07106 \dots \cdot \int \frac{1}{v ^{2} - 36} d v

Aplicar integración por sustitución trigonométrica

: \int sec (w) d w

= 7.07106 \dots \cdot \int sec (w) d w

Aplicar la regla de integración:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= 7.07106 \dots ln ∣ tan (w) + sec (w) ∣

Sustitución hacia atrás

= 7.07106 \dots ln tan (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) + sec (arcsec (\frac{1}{6} (0.02 x + 12 - 6)))

Simplificar

7.07106 \dots ln tan (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) + sec (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) : 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6)

= 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6)

Agregar una constante a la solución

= 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6) + C

\frac{d}{d x} (sin (ln ((x^{2} + 4)^{4})))

7 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

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

x \to 8 lim (\frac{x ^{2} - 5}{x - 8})

\int x (3 x^{2} + 1) d x