| $\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \frac{d}{{dx}}\left( {\ln x} \right)=\frac{1}{x}} & {\displaystyle \int{{\frac{1}{x}}}\ dx=\ln \left| x \right|+C} \\ \hline {\displaystyle \frac{d}{{dx}}\left[ {\ln \left( {ax+b} \right)} \right]=\frac{a}{{ax+b}}} & {\displaystyle \int{{\frac{1}{{ax+b}}}}\ dx=\frac{1}{a}\ln \left| {ax+b} \right|+C} \\ \hline { \displaystyle \frac{d}{{dx}}\left( {{{e}^{x}}} \right)={{e}^{x}}} & {\displaystyle \int{{{{e}^{x}}}}\ dx={{e}^{x}}+C}\\ \hline {\displaystyle \frac{d}{{dx}}\left( {{{e}^{{ax+b}}}} \right)=a{{e}^{{ax+b}}}} & {\displaystyle \int{{{{e}^{{ax+b}}}}}\ dx=\frac{1}{a}{{e}^{{ax+b}}}+C} \\ \hline \end{array}\end{array}$ |
1. Integrate each of the following with respect to $ \displaystyle x$
$ \displaystyle \begin{array}{l}\text{(a)}\ \ \displaystyle \frac{1}{{3x}}\\\\\text{(b)}\ \ \displaystyle \frac{2}{{5x}}\\\\\text{(c)}\ \ \displaystyle \frac{3}{{4x-1}}\\\\\text{(d)}\ \ \displaystyle \frac{{10}}{{25x+3}}\\\\\text{(e)}\ \ \displaystyle \frac{7}{{2-5x}}\end{array}$
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| $ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{1}{{3x}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{3}\displaystyle \int{{\displaystyle \frac{1}{x}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{3}\ln |x|+C\\\\\\\\\text{(b)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{2}{{5x}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{5}\displaystyle \int{{\displaystyle \frac{1}{x}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{5}\ln |x|+C\\\\\\\\\text{(c)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{3}{{4x-1}}}}\ dx\\\\\ \ \ \ =\ \ 3\displaystyle \int{{\displaystyle \frac{1}{{4x-1}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{3}{4}\ln |4x-1|+C\\\\\\\\\text{(d)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{10}}{{25x+3}}}}\ dx\\\\\ \ \ \ =\ \ 10\displaystyle \int{{\displaystyle \frac{1}{{25x+3}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{{10}}{{25}}\ln |25x+3|+C\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{5}\ln |25x+3|+C\\\\\\\\\text{(e)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{7}{{2-5x}}}}\ dx\\\\\ \ \ \ =\ \ 7\displaystyle \int{{\displaystyle \frac{1}{{2-5x}}}}\ dx\\\\\ \ \ \ =\ \ -\displaystyle \frac{7}{5}\ln |2-5x|+C\end{array}$ |
2. Find each of the following indefinite integrals.
$ \displaystyle \begin{array}{l}\text{(a)}\ \ \displaystyle \int{{{{e}^{{3x}}}}}\ dx\\\\\text{(b)}\ \ \displaystyle \int{{{{e}^{{2-5x}}}}}\ dx\\\\\text{(c)}\ \ \displaystyle \int{{{{e}^{{3x+\pi }}}}}\ dx\\\\\text{(d)}\ \ \displaystyle \int{{{{{\left( {{{e}^{{2x}}}+1} \right)}}^{2}}}}\ dx\end{array}$
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| $ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \displaystyle \int{{{{e}^{{3x}}}}}\ dx\\\\\ \ \ \ =\displaystyle \frac{1}{3}{{e}^{{3x}}}+C\\\\\\\\\text{(b)}\ \ \ \displaystyle \int{{{{e}^{{2-5x}}}}}\ dx\\\\\ \ \ \ =-\displaystyle \frac{1}{5}{{e}^{{2-5x}}}+C\\\\\\\\\text{(c)}\ \ \ \displaystyle \int{{{{e}^{{3x+\pi }}}}}\ dx\\\\\ \ \ \ =\displaystyle \frac{1}{3}{{e}^{{3x+\pi }}}+C\\\\\\\\\text{(d)}\ \ \ \displaystyle \int{{{{{\left( {{{e}^{{2x}}}+1} \right)}}^{2}}}}\ dx\\\\\ \ \ \ =\ \displaystyle \int{{\left( {{{e}^{{4x}}}+2{{e}^{{2x}}}+1} \right)}}\ dx\\\\\ \ \ \ =\ \displaystyle \int{{{{e}^{{4x}}}}}\ dx+2\displaystyle \int{{{{e}^{{2x}}}}}\ dx+\displaystyle \int{1}\ dx\\\\\ \ \ \ =\ \displaystyle \frac{1}{4}{{e}^{{4x}}}+{{e}^{{2x}}}+x+C\end{array}$ |
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