Question

Calcular I = ∫ t (t^2 + 1) dt

Answer 1

Hola!

Respuesta:

$I = \frac{ {( {t}^{2} + 1)}^{2} }{4} + c$

Explicación paso a paso:

$I = \int \: t( {t}^{2} + 1) \: \: \: dt$

∆) Aplicando cambio de variable:

$u = {t}^{2} + 1$

$du = 2t \: dt$

$\frac{du}{2} = t \: dt$

∆) Reemplazamos:

$I = \int \: t( {t}^{2} + 1) \: \: \: dt$

$I = \int \: (u) \: \: \: \frac{du}{2}$

$I = \frac{1}{2} \int \: (u) \: \: \: du$

$I = \frac{1}{2} \: \: \frac{ {u}^{2} }{2} + c$

$I = \frac{ {u}^{2} }{4} + c$

$I = \frac{ {( {t}^{2} + 1)}^{2} }{4} + c$