a) Cách 1: Đặt \(u = 1 - x \Rightarrow du= -dx\). Khi đó ta được \(-\int u^{9}du = -\dfrac{1}{10}u^{10}+C\)
Suy ra \(\int(1-x)^{9}dx=-\dfrac{(1-x)^{10}}{10}+C\)
Cách 2: \(\smallint {\left( {1 - x} \right)^9}dx = - \smallint {\left( {1 - x} \right)^{9}}d\left( {1 - x} \right)=\) \(-\dfrac{(1-x)^{10}}{10} +C\)
\(b)\;\;\int {x{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}dx} .\)
Cách 1: Đặt \(u = 1 + {x^2} \Rightarrow du = 2xdx \Rightarrow xdx = \dfrac{1}{2}du.\) \( \Rightarrow \int {\dfrac{1}{2}{u^{\dfrac{3}{2}}}du =\dfrac{1}{2}.\dfrac{{{u^{\dfrac{3}{2} + 1}}}}{{\dfrac{3}{2} + 1}} + C = \dfrac{{{u^{\dfrac{5}{2}}}}}{5} + C =\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{5}{2}}}}}{5}} +C.\)
Cách 2: \(\int x(1+x^{2})^{\dfrac{3}{2}}dx= \dfrac{1}{2}\int (1+x^{2})^{\dfrac{3}{2}}d(1+x^2{}) \\= \dfrac{1}{2}.\dfrac{2}{5}(1+x^{2})^{\dfrac{5}{2}}+C = \dfrac{1}{5}.(1+x^{2})^{\dfrac{5}{2}}+C\)
\(c)\;\;{\cos ^3}x.\sin xdx.\)
Cách 1: Đặt: \(t = {\mathop{\rm cosx}\nolimits} \Rightarrow dt = - sinxdx.\)
\(\begin{array}{l} \Rightarrow \int {{{\cos }^3}x.{\mathop{\rm sinxdx}\nolimits} } = \int { - {t^3}du} \\ = - \dfrac{1}{4}{t^4} + C = - \dfrac{1}{4}{\cos ^4}x + C.\end{array}\)
Cách 2: \(∫cos^3xsinxdx = -∫cos^3xd(cosx)\\= -\dfrac{1}{4}.cos^{4}x + C.\)
\(d)\;\;\int {\dfrac{{dx}}{{{e^x} + {e^{ - x}} + 2}}.} \)
Cách 1:
Ta có: \({e^x} + {e^{ - x}} + 2 = {e^x} + \dfrac{1}{{{e^x}}} + 2 = \dfrac{{{e^{2x}} + 2{e^x} + 1}}{{{e^x}}} = \dfrac{{{{\left( {{e^x} + 1} \right)}^2}}}{{{e^x}}}.\)
\( \Rightarrow \dfrac{1}{{{e^x} + {e^{ - x}} + 2}} = \dfrac{{{e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}}.\)
Đặt \(u = {e^x} + 1 \Rightarrow du = {e^x}dx.\)
\(\int {\dfrac{{dx}}{{{e^x} + {e^{ - x}} + 2}}} = \int {\dfrac{{{e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}}dx} \) \( = \int {\dfrac{{du}}{{{u^2}}}} = - \dfrac{1}{u} + C = - \dfrac{1}{{{e^x} + 1}} + C\)
Cách 2: \(\int \dfrac{dx}{e^{x}+e^{-x}+2} = \int \dfrac{e^{x}}{e^{2x}+2e^{x}+1}dx\\ = \int \dfrac{d(e^{x}+1)}{(e^{x}+1)^{2}}dx=\dfrac{-1}{e^{x}+1} + C.\)
