integral de (sqrt(x^2-25))/(5x)

Solución

\int \frac{x ^{2} - 25}{5 x} d x

- arcsec (\frac{1}{5} x) + \frac{1}{5} x^{2} - 25 + C

Pasos de solución

\int \frac{x ^{2} - 25}{5 x} d x

Sacar la constante:

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

= \frac{1}{5} \cdot \int \frac{x ^{2} - 25}{x} d x

Aplicar integración por sustitución trigonométrica

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

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

Sacar la constante:

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

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

Re-escribir usando identidades trigonométricas

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

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}{5} \cdot 5 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= \frac{1}{5} \cdot 5 (- u + tan (u))

Sustituir en la ecuación

u = arcsec (\frac{1}{5} x)

= \frac{1}{5} \cdot 5 (- arcsec (\frac{1}{5} x) + tan (arcsec (\frac{1}{5} x)))

Simplificar

\frac{1}{5} \cdot 5 (- arcsec (\frac{1}{5} x) + tan (arcsec (\frac{1}{5} x))) : - arcsec (\frac{1}{5} x) + \frac{1}{5} x^{2} - 25

= - arcsec (\frac{1}{5} x) + \frac{1}{5} x^{2} - 25

Agregar una constante a la solución

= - arcsec (\frac{1}{5} x) + \frac{1}{5} x^{2} - 25 + C

\int x^{3} (x^{4} - 5)^{4} d x

\int 3 x - \frac{sin ( x )}{cos ^{2} ( x )} d x

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

\int \frac{x ^{3} - 3 x - 8}{x - 4} d x

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