Question

encontrar la
integral inmediata ∫(t^3-1)/(2t-2) dt

Answer 1

Respuesta:

$\frac{1}{6}t^3+\frac{1}{4}t^2+\frac{1}{2}t+C$

Explicación paso a paso:

Podes factorizar:

• $t^3-1=(t-1)(t^2+t+1)$
• $2t-2=2(t-1)$

Entonces:

$\int\frac{t^3-1}{2t-2}\,dt=\\\\=\int\frac{(t-1)(t^2+t+1)}{2(t-1)}\,dt\\\\=\frac{1}{2}\int(t^2 +t+1)\,dt\\\\=\frac{1}{2}(\frac{1}{3}t^3+\frac{1}{2}t^2+t)+C\\\\=\frac{1}{6}t^3+\frac{1}{4}t^2+\frac{1}{2}t+C$