a) Điều kiện \(x>0\). Thực hiện chia tử cho mẫu ta được:
\(f(x) = \dfrac{x+x^{\frac{1}{2}}+1}{x^{\frac{1}{3}}} \\= x^{1-\frac{1}{3}}+ x^{\frac{1}{2}-\frac{1}{3}}+ x^{-\frac{1}{3}}\\ = x^{\frac{2}{3}}+ x^{\frac{1}{6}} + x^{-\frac{1}{3}}.\)
\(\Rightarrow ∫f(x)dx = ∫(x^{\frac{2}{3}}+ x^{\frac{1}{6}} + x^{-\frac{1}{3}})dx \\= \dfrac{3}{5}x^{\frac{5}{3}}+ \dfrac{6}{7}x^{\frac{7}{6}}+\dfrac{3}{2}x^{\frac{2}{3}} +C.\)
\(\begin{array}{l}b)\;\;f\left( x \right) = \dfrac{{{2^x} - 1}}{{{e^x}}} = {\left( {\dfrac{2}{e}} \right)^x} - {e^{ - x}}.\\ \Rightarrow F\left( x \right) = \int {f\left( x \right)dx} = \int {\left( {{{\left( {\dfrac{2}{e}} \right)}^x} - {e^{ - x}}} \right)} dx\\= \dfrac{{{{\left( {\dfrac{2}{e}} \right)}^x}}}{{\ln \left( {\dfrac{2}{e}} \right)}} + {e^{ - x}} + C = \dfrac{{{2^x}}}{{{e^x}\left( {\ln 2 - 1} \right)}} + \dfrac{1}{{{e^x}}} + C\\= \dfrac{{{2^x} + \ln 2 - 1}}{{{e^x}\left( {\ln 2 - 1} \right)}} + C.\end{array}\)
\(\begin{array}{l}c)\;\;f\left( x \right) = \dfrac{1}{{{{\sin }^2}x.{{\cos }^2}x}} = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} \\= \dfrac{{{{\sin }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} = \dfrac{1}{{{{\sin }^2}x}} + \dfrac{1}{{{{\cos }^2}x}}.\\\Rightarrow F\left( x \right) = \int {f\left( x \right)dx = \int {\left( {\dfrac{1}{{{{\sin }^2}x}} + \dfrac{1}{{{{\cos }^2}x}}} \right)} } dx\\ = - \cot x + \tan x + C = \dfrac{{\sin x}}{{\cos x}} - \dfrac{{\cos x}}{{\sin x}} + C\\ = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x.\cos x}} + C = \dfrac{{ - \cos 2x}}{{\dfrac{1}{2}\sin 2x}} + C = - 2\cot2 x + C.\end{array}\)
d) Áp dụng công thức biến đổi tích thành tổng ta có:
\(\begin{array}{l}f\left( x \right) = \sin 5x.\cos 3x = \dfrac{1}{2}\left( {\sin 8x + \sin 2x} \right).\\\Rightarrow F\left( x \right) = \int {f\left( x \right)dx} = \int {\dfrac{1}{2}\left( {\sin 8x + \sin 2x} \right)dx} \\ = \dfrac{1}{2}\left( { - \dfrac{1}{8}\cos 8x - \dfrac{1}{2}\cos 2x} \right) + C\\ = - \dfrac{1}{4}\left( {\dfrac{1}{4}\cos 8x + \cos 2x} \right) + C.\end{array}\)
\(\begin{array}{l}e)\;\;f\left( x \right) = {\tan ^2}x = \dfrac{1}{{{{\cos }^2}x}} - 1\\\Rightarrow F\left( x \right) = \int {f\left( x \right)dx} = \int {\left( {\dfrac{1}{{{{\cos }^2}x}} - 1} \right)dx}\\ = \tan x - x + C.\end{array}\)
\(\begin{array}{l}g)\;\;f\left( x \right) = {e^{3 - 2x}}.\\\Rightarrow F\left( x \right) = \int {f\left( x \right)dx = } \int {{e^{3 - 2x}}dx} \\= - \dfrac{1}{2}\int {{e^{3 - 2x}}\left( {3 - 2x} \right)'dx} = - \dfrac{1}{2}{e^{3 - 2x}} + C.\end{array}\)
h) Ta có : \(f\left( x \right) = \dfrac{1}{{\left( {1 + x} \right)\left( {1 - 2x} \right)}} = \dfrac{1}{{3\left( {x + 1} \right)}} + \dfrac{2}{{3\left( {1 - 2x} \right)}}.\)
\(\Rightarrow \int \dfrac{dx}{(1+x)(1-2x)}=\dfrac{1}{3}\int (\dfrac{1}{1+x}+\dfrac{2}{1-2x})dx \\= \dfrac{1}{3}(ln\left | 1+x \right |)-ln\left | 1-2x \right |)+C\\ = \dfrac{1}{3}ln\left | \dfrac{1+x}{1-2x} \right | +C.\)
