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integral de (t^5)/(sqrt(t^2+7))

Solución

\int \frac{t ^{5}}{t ^{2} + 7} d t

497 \frac{1}{7} (7 + t^{2}) - \frac{14}{3} (t^{2} + 7)^{\frac{3}{2}} + \frac{1}{5} (7 + t^{2})^{\frac{5}{2}} + C

Pasos de solución

\int \frac{t ^{5}}{t ^{2} + 7} d t

Aplicar integración por sustitución trigonométrica

: \int 497 tan^{5} (u) sec (u) d u

= \int 497 tan^{5} (u) sec (u) d u

Sacar la constante:

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

= 497 \cdot \int tan^{5} (u) sec (u) d u

Re-escribir usando identidades trigonométricas

= 497 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) sec (u) d u

Aplicar integración por sustitución

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

= 497 \cdot \int (- 1 + v^{2})^{2} d v

Expandir

(- 1 + v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= 497 \cdot \int 1 - 2 v^{2} + v^{4} 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

= 497 (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= 497 (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Sustitución hacia atrás

= 497 sec (arctan (\frac{1}{7} t)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{7} t ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{7} t ) )}{5}

Simplificar

497 sec (arctan (\frac{1}{7} t)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{7} t ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{7} t ) )}{5} : 497 \frac{1}{7} (7 + t^{2}) - \frac{14}{3} (t^{2} + 7)^{\frac{3}{2}} + \frac{1}{5} (7 + t^{2})^{\frac{5}{2}}

= 497 \frac{1}{7} (7 + t^{2}) - \frac{14}{3} (t^{2} + 7)^{\frac{3}{2}} + \frac{1}{5} (7 + t^{2})^{\frac{5}{2}}

Agregar una constante a la solución

= 497 \frac{1}{7} (7 + t^{2}) - \frac{14}{3} (t^{2} + 7)^{\frac{3}{2}} + \frac{1}{5} (7 + t^{2})^{\frac{5}{2}} + C

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