Đáp án:
$\begin{array}{l}
a)\int {\frac{{dx}}{{1 + \sqrt[3]{{x + 1}}}}} \\
Đặt:\sqrt[3]{{x + 1}} = u\\
\Rightarrow x + 1 = {u^3}\\
\Rightarrow dx = 3{u^2}du\\
\int {\frac{{dx}}{{1 + \sqrt[3]{{x + 1}}}}} = \int {\frac{{3{u^2}du}}{{1 + u}}} \\
= \int {3u - 3 + \frac{3}{{u + 1}}du} \\
= \frac{3}{2}{u^2} - 3u + 3\ln \left( {u + 1} \right) + C\\
= \frac{3}{2}\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} - 3\sqrt[3]{{x + 1}} + 3.\ln \left( {\sqrt[3]{{x + 1}}} \right) + C\\
b)\int {{{\left( {1 - t{\rm{anx}}} \right)}^4}\frac{1}{{co{s^2}x}}} dx\\
= \int {{{\left( {\tan x - 1} \right)}^4}.d\left( {\tan x - 1} \right)} \\
= \frac{1}{5}.{\left( {\tan x - 1} \right)^5} + C
\end{array}$
