Giải thích các bước giải:
$\begin{array}{l}
a)f\left( x \right) = \dfrac{x}{{\sqrt {1 + x} }}\\
\int {f\left( x \right)dx} = \int {\dfrac{x}{{\sqrt {1 + x} }}dx} \\
= \int {2x.\dfrac{1}{{2\sqrt {1 + x} }}dx} \\
= \int {2xd\left( {\sqrt {1 + x} } \right)} \\
= 2x\sqrt {1 + x} - \int {2\sqrt {1 + x} } dx\\
= 2x\sqrt {1 + x} - 2\int {{{\left( {1 + x} \right)}^{\dfrac{1}{2}}}dx} \\
= 2x\sqrt {1 + x} - 2.\dfrac{2}{3}{\left( {1 + x} \right)^{\dfrac{3}{2}}} + C\\
= 2x\sqrt {1 + x} - \dfrac{4}{3}\left( {1 + x} \right)\sqrt {1 + x} + C\\
b)f\left( x \right) = \sqrt {1 + \sin 2x} \\
\int {f\left( x \right)dx} = \int {\sqrt {1 + \sin 2x} dx} \\
= \int {\sqrt {{{\left( {\sin x + \cos x} \right)}^2}} dx} \\
= \int {\left| {\sin x + \cos x} \right|dx} \\
= \int {\sqrt 2 \left| {\sin \left( {x + \dfrac{\pi }{4}} \right)} \right|dx} \\
= \sqrt 2 \int {\left| {\sin \left( {x + \dfrac{\pi }{4}} \right)} \right|dx} \\
= \left\{ \begin{array}{l}
\sqrt 2 \int {\sin \left( {x + \dfrac{\pi }{4}} \right)dx} ,0 \le \sin \left( {x + \dfrac{\pi }{4}} \right) \le 1\\
\sqrt 2 \int { - \sin \left( {x + \dfrac{\pi }{4}} \right)dx, - 1 \le \sin \left( {x + \dfrac{\pi }{4}} \right) < 0}
\end{array} \right.\\
= \left\{ \begin{array}{l}
- \sqrt 2 \cos \left( {x + \dfrac{\pi }{4}} \right) + C,0 \le \sin \left( {x + \dfrac{\pi }{4}} \right) \le 1\\
\sqrt 2 \cos \left( {x + \dfrac{\pi }{4}} \right) + C, - 1 \le \sin \left( {x + \dfrac{\pi }{4}} \right) < 0
\end{array} \right.
\end{array}$
