# Integration by Substitution

 $\displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{f(g(x)){g}'(x)}}\ dx\\\\\,\ \ \ \text{Let}\ u=g(x).\\\\\therefore \ \ \displaystyle \frac{{du}}{{dx}}={g}'(x)\\\\\therefore \ \ du={g}'(x)dx\end{array}$ $\displaystyle \begin{array}{l}\begin{array}{|r|l|l|l|c|c|c|c|c|c|} \hline { \displaystyle \therefore \int{{f(g(x)){g}'(x)}}\ dx=\int{{f(u)}}\ du} \\ \hline \end{array}\end{array}$

##### 1.       Evaluate

$\displaystyle \text{(a)}\ \ \ \ \int{{(2{{x}^{3}}-3x+1)(2x-1)}}\ dx$

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 $\displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{(2{{x}^{3}}-3x+1)(2x-1)}}\ dx\\\\\,\ \ \ \text{Let}\ u=2{{x}^{3}}-3x+1.\\\\\therefore \ \ du=(6x-3)dx\\\\\therefore \ \ du=3(2x-1)dx\\\\\therefore \ \ (2x-1)dx=\displaystyle \frac{1}{3}du\\\\\therefore \ \ \displaystyle \int{{(2{{x}^{3}}-3x+1)(2x-1)}}\ dx\\\\\ =\ \displaystyle \int{{u\cdot }}\displaystyle \frac{1}{3}du\\\\\ =\ \displaystyle \frac{1}{3}\displaystyle \int{{u\ }}du\\\\\ =\ \displaystyle \frac{1}{6}{{u}^{2}}+C\\\\\ =\ \displaystyle \frac{1}{6}{{\left( {2{{x}^{3}}-3x+1} \right)}^{2}}+C\end{array}$

$\displaystyle \text{(b)}\ \ \ \ \int{{\frac{{{{{\left( {2-\sqrt{x}} \right)}}^{5}}}}{{\sqrt{x}}}}}\ dx$

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 $\displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{{\left( {2-\sqrt{x}} \right)}}^{5}}}}{{\sqrt{x}}}}}\ dx\\\\\,\ \ \ \text{Let}\ u=2-\sqrt{x}.\\\\\therefore \ \ du=-\displaystyle \frac{1}{{2\sqrt{x}}}dx\\\\\therefore \ \ \displaystyle \frac{1}{{\sqrt{x}}}dx=-2du\\\\\therefore \ \ \displaystyle \int{{\displaystyle \frac{{{{{\left( {2-\sqrt{x}} \right)}}^{5}}}}{{\sqrt{x}}}}}\ dx\\\\\ =\ -2\displaystyle \int{{{{u}^{5}}\ }}du\\\\\ =\ -2\left( {\displaystyle \frac{1}{6}{{u}^{6}}} \right)+C\\\\\ =\ -\displaystyle \frac{1}{3}{{u}^{6}}+C\\\\\ =\ -\displaystyle \frac{1}{3}{{\left( {2-\sqrt{x}} \right)}^{6}}+C\end{array}$

$\displaystyle (\text{c})\ \ \ \ \int{{\frac{{{{x}^{2}}}}{{5{{x}^{3}}-2}}}}\ dx$

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 $\displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}}}{{5{{x}^{3}}-2}}}}\ dx\\\\\,\ \ \ \text{Let}\ u=5{{x}^{3}}-2.\\\\\therefore \ \ du=15{{x}^{2}}\ dx\\\\\therefore \ \ {{x}^{2}}\ dx=\displaystyle \frac{1}{{15}}du\\\\\therefore \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}}}{{5{{x}^{3}}-2}}}}\ dx\\\\\ =\ \displaystyle \frac{1}{{15}}\displaystyle \int{{\displaystyle \frac{1}{u}\ }}du\\\\\ =\ \displaystyle \frac{1}{{15}}\ln |u|+C\\\\\ =\ \displaystyle \frac{1}{{15}}\ln |5{{x}^{3}}-2|+C\end{array}$

$\displaystyle (\text{d)}\ \ \ \ \int{{\frac{{{{e}^{{3x}}}}}{{1+{{e}^{x}}}}}}\ dx$

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 $\displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{e}^{{3x}}}}}{{1+{{e}^{x}}}}}}\ dx\\\\=\ \ \displaystyle \int{{\displaystyle \frac{{{{e}^{{2x}}}\cdot {{e}^{x}}}}{{1+{{e}^{x}}}}}}\ dx\\\\=\ \ \displaystyle \int{{\displaystyle \frac{{{{{\left( {{{e}^{x}}} \right)}}^{2}}\cdot {{e}^{x}}}}{{1+{{e}^{x}}}}}}\ dx\\\\\,\ \ \ \text{Let}\ u=1+{{e}^{x}},\ \text{then}\ {{e}^{x}}=u-1\\\\\therefore \ \ du={{e}^{x}}\ dx\\\\\therefore \ \ \displaystyle \int{{\displaystyle \frac{{{{e}^{{3x}}}}}{{1+{{e}^{x}}}}}}\ dx\\\\\ =\ \displaystyle \int{{\displaystyle \frac{{{{{\left( {u-1} \right)}}^{2}}}}{u}}}\ du\\\\\ =\ \displaystyle \int{{\displaystyle \frac{{{{u}^{2}}-2u+1}}{u}}}\ du\\\\\ =\ \displaystyle \int{{\left( {u-2+\displaystyle \frac{1}{u}} \right)}}\ du\\\\=\ \displaystyle \int{u}\ du-2\displaystyle \int{{du}}\ +\displaystyle \int{{\displaystyle \frac{1}{u}}}\ du\\\\\ =\ \displaystyle \frac{1}{2}{{u}^{2}}-2u+\ln |u|+C\\\\\ =\ \displaystyle \frac{1}{2}{{\left( {1+{{e}^{x}}} \right)}^{2}}-2\left( {1+{{e}^{x}}} \right)+\ln |1+{{e}^{x}}|+C\\\\\ =\ \displaystyle \frac{1}{2}\left( {1+2{{e}^{x}}+{{e}^{{2x}}}} \right)-2\left( {1+{{e}^{x}}} \right)+\ln |1+{{e}^{x}}|+C\\\\\ =\ \displaystyle \frac{1}{2}+{{e}^{x}}+\displaystyle \frac{1}{2}{{e}^{{2x}}}-2-2{{e}^{x}}+\ln |1+{{e}^{x}}|+C\\\\\ =\ \displaystyle \frac{1}{2}{{e}^{{2x}}}-{{e}^{x}}+\ln |1+{{e}^{x}}|-\displaystyle \frac{3}{2}+C\\\\\ =\ \displaystyle \frac{1}{2}{{e}^{{2x}}}-{{e}^{x}}+\ln |1+{{e}^{x}}|+\ k\ \ \ \left( {k=C-\displaystyle \frac{3}{2}} \right)\end{array}$

$\displaystyle (\text{e})\ \ \ \ \int{{\frac{{{{{(\ln x)}}^{8}}+1}}{x}}}\ dx$

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 $\displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{{(\ln x)}}^{8}}+1}}{x}}}\ dx\\\\\,\ \ \ \text{Let}\ u=\ln x,\ \\\\\therefore \ \ du=\displaystyle \frac{1}{x}\ dx\\\\\therefore \ \ \displaystyle \int{{\displaystyle \frac{{{{{(\ln x)}}^{8}}+1}}{x}}}\ dx\\\\\therefore \ \ \displaystyle \int{{\left[ {{{{(\ln x)}}^{8}}+1} \right]\displaystyle \frac{1}{x}}}\ dx\\\\\ =\ \displaystyle \int{{\left( {{{u}^{8}}+1} \right)}}\ du\\\\=\ \displaystyle \int{{{{u}^{8}}}}\ du+2\displaystyle \int{{du}}\ \\\\\ =\ \displaystyle \frac{1}{9}{{u}^{9}}+2u+C\\\\\ =\ \displaystyle \frac{1}{9}{{\left( {\ln x} \right)}^{9}}+2\left( {\ln x} \right)+C\end{array}$

$\displaystyle \text{(f)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}-8x+3}}{{x+2}}}}\ dx$

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 $\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}-8x+3}}{{x+2}}}}\ dx\\\\\ \ \ =\ \ \displaystyle \int{{\displaystyle \frac{{{{{\left( {x+2} \right)}}^{2}}-12(x+2)+23}}{{x+2}}}}\ dx\\\\\ \ \ \ \ \ \ \text{Let}\ u=x+2\\\\\therefore \ \ \ \ \ du=dx\\\\\therefore \ \ \ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}-8x+3}}{{x+2}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\displaystyle \frac{{{{u}^{2}}-12u+23}}{u}}}\ du\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {u-12+\displaystyle \frac{{23}}{u}} \right)}}\ du\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {u-12+\displaystyle \frac{{23}}{u}} \right)}}\ du\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{2}{{u}^{2}}-12u+23\ln |u|+C\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{2}{{\left( {x+2} \right)}^{2}}-12\left( {x+2} \right)+23\ln |x+2|+C\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{2}{{x}^{2}}-10x+23\ln |x+2|-22+C\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{2}{{x}^{2}}-10x+23\ln |x+2|+k\ \left[ {k=C-22} \right]\end{array}$

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