integral de sqrt(x^2-4x+3)

Solución

\int x^{2} - 4 x + 3 d x

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

Pasos de solución

\int x^{2} - 4 x + 3 d x

Completar el cuadrado

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

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

Aplicar integración por sustitución

: \int u^{2} - 1 d u

= \int u^{2} - 1 d u

Aplicar integración por sustitución trigonométrica

: \int tan^{2} (v) sec (v) d v

= \int tan^{2} (v) sec (v) d v

Re-escribir usando identidades trigonométricas

= \int (- 1 + sec^{2} (v)) sec (v) d v

Expandir

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= \int - sec (v) + sec^{3} (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

= - \int sec (v) d v + \int sec^{3} (v) d v

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

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= - ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Sustitución hacia atrás

= - ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣ + \frac{1}{2} sec (arcsec (x - 2)) tan (arcsec (x - 2)) + \frac{1}{2} ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣

Simplificar

- ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣ + \frac{1}{2} sec (arcsec (x - 2)) tan (arcsec (x - 2)) + \frac{1}{2} ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣ : - ln x^{2} - 4 x + 3 + x - 2 + \frac{1}{2} (x - 2) x^{2} - 4 x + 3 + \frac{1}{2} ln x^{2} - 4 x + 3 + x - 2

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

Agregar una constante a la solución

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

\int \frac{y}{3 y + 1} d y

\frac{d y}{d x} = 3 cos (x) y, y (\frac{π}{2}) = - 2

\frac{d}{d x} ((x^{2}) (\frac{2}{x ^{2}} + \frac{5}{x ^{3}}))

x \to - \infty lim (0. 1^{x})

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