Tính \( I = \smallint 3{x^5}\sqrt {{x^3} + 1} dx\)

\(
I = \smallint 3{x^5}\sqrt {{x^3} + 1} dx = \smallint 3{x^2}.{x^3}\sqrt {{x^3} + 1} dx\)

Đặt

\(\begin{array}{l}
\sqrt {{x^3} + 1} = t \Rightarrow {x^3} + 1 = {t^2} \Rightarrow 3{x^2}dx = 2tdt\\
\Rightarrow I = \smallint \left( {{t^2} – 1} \right).t.2tdt = 2\smallint \left( {{t^4} – {t^2}} \right)dt = \frac{2}{5}{t^5} – \frac{2}{3}{t^3} + C\\
= \frac{2}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1} – \frac{2}{3}\left( {{x^3} + 1} \right)\sqrt {{x^3} + 1} + C
\end{array}\)