Giải thích các bước giải:
$I=\int e^{\sqrt{7x+4}}dx$
Đặt $\sqrt{7x+4}=u\to \dfrac{7}{2\sqrt{7x+4}}dx=du\to \dfrac{7}{2u}dx=du\to dx=\dfrac{2u}{7}du$
$\to I=\int \dfrac{2e^uu}{7}du$
$\to I=\dfrac{2}{7}\int ud(e^u)$
$\to I=\dfrac{2}{7}(ue^u-\int e^udu)$
$\to I=\dfrac{2}{7}(ue^u-e^u)+C$
$\to I=\dfrac{2}{7}\left(e^{\sqrt{7x+4}}\sqrt{7x+4}-e^{\sqrt{7x+4}}\right)+C$
