a) Đặt \(t = 1 + {x^3}\)\( \Rightarrow dt = 3{x^2}dx \Rightarrow {x^2}dx = \dfrac{{dt}}{3}\).
Khi đó \(\int {{x^2}\sqrt[3]{{1 + {x^3}}}} dx = \int {\sqrt[3]{t}.\dfrac{{dt}}{3}} \) \( = \dfrac{1}{3}\int {{t^{\dfrac{1}{3}}}dt} = \dfrac{1}{3}.\dfrac{{{t^{\dfrac{1}{3} + 1}}}}{{\dfrac{1}{3} + 1}} + C\) \( = \dfrac{1}{4}{t^{\dfrac{4}{3}}} + C = \dfrac{1}{4}{\left( {1 + {x^3}} \right)^{\dfrac{4}{3}}} + C\)
b) Đặt \(t = {x^2} \Rightarrow dt = 2xdx\) \( \Rightarrow xdx = \dfrac{{dt}}{2}\)
Khi đó \(\int {x{e^{ - {x^2}}}} dx = \int {{e^{ - t}}.\dfrac{{dt}}{2}} \)\( = - \dfrac{1}{2}{e^{ - t}} + C = - \dfrac{1}{2}{e^{ - {x^2}}} + C\).
c) Đặt \(t = 1 + {x^2}\)\( \Rightarrow dt = 2xdx \Rightarrow xdx = \dfrac{{dt}}{2}\).
Khi đó, \(\int {\dfrac{x}{{{{(1 + {x^2})}^2}}}} dx = \int {\dfrac{1}{{{t^2}}}.\dfrac{{dt}}{2}} = \dfrac{1}{2}\int {\dfrac{{dt}}{{{t^2}}}} \) \( = - \dfrac{1}{2}.\dfrac{1}{t} + C = - \dfrac{1}{{2\left( {1 + {x^2}} \right)}} + C\)
d) Đặt \(t = \sqrt x \Rightarrow dt = \dfrac{1}{{2\sqrt x }}dx\)\( \Rightarrow \dfrac{{dx}}{{\sqrt x }} = 2dt\) và \(x = {t^2}\).
Khi đó \(\int {\dfrac{1}{{(1 - x)\sqrt x }}} dx\)\( = \int {\dfrac{1}{{\left( {1 - {t^2}} \right)}}.2dt} = \int {\dfrac{2}{{1 - {t^2}}}dt} \) \( = \int {\left( {\dfrac{1}{{1 - t}} + \dfrac{1}{{1 + t}}} \right)dt} \)
\( = - \ln \left| {1 - t} \right| + \ln \left| {1 + t} \right| + C\) \( = \ln \left| {\dfrac{{1 + t}}{{1 - t}}} \right| + C\)\( = \ln \left| {\dfrac{{1 + \sqrt x }}{{1 - \sqrt x }}} \right| + C\).
e) Đặt \(t = \dfrac{1}{x}\)\( \Rightarrow dt = - \dfrac{1}{{{x^2}}}dx \Rightarrow \dfrac{{dx}}{{{x^2}}} = - dt\).
Khi đó \(\int {\sin \dfrac{1}{x}.\dfrac{1}{{{x^2}}}} dx\)\( = \int {\sin t.\left( { - dt} \right)} = \int {\left( { - \sin t} \right)dt} \) \( = \cos t + C = \cos \dfrac{1}{x} + C\)
g) Đặt \(t = \ln x\)\( \Rightarrow dt = \dfrac{{dx}}{x}\). Khi đó
\(\int {\dfrac{{{{(\ln x)}^2}}}{x}} dx = \int {{t^2}.dt} \)\( = \dfrac{{{t^3}}}{3} + C = \dfrac{{{{\ln }^3}x}}{3} + C\)
h) Đặt \(t = \cos x\)\( \Rightarrow dt = - \sin xdx\).
Khi đó \(\int {\dfrac{{\sin x}}{{\sqrt[3]{{{{\cos }^2}x}}}}} dx\)\( = \int {\dfrac{{ - dt}}{{\sqrt[3]{{{t^2}}}}}} = \int { - {t^{ - \dfrac{2}{3}}}dt} \) \( = - \dfrac{{{t^{ - \dfrac{2}{3} + 1}}}}{{ - \dfrac{2}{3} + 1}} + C = - 3{t^{\dfrac{1}{3}}} + C\) \( = - 3\sqrt[3]{t} + C = - 3\sqrt[3]{{\cos x}} + C\).
