5.5 The Substitution Rule/13: Difference between revisions
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(Created page with "<math> \int \frac{\sin{(\ln{(x))}}}{x}dx </math> <math> \begin{align} u &=\ln(x) \\[2ex] du &= \frac{1}{x}dx \\[2ex] \end{align} </math> <math> \begin{align} \int \frac{\sin{(\ln{(x))}}}{x}dx &= \int\frac{1}{x}\sin(\ln{(x)})dx = \int\left(\frac{1}{x}dx\right)\sin{(\ln{(x)})} \\[2ex] &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] &= -\cos{(u)} + C \\[2ex] &= -\cos{(\ln{(x)})} + C \end{align} </math>") |
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<math> | <math> | ||
\int \frac{ | \int \frac{dx}{5-3x} \text{,} \quad u=5-3x | ||
</math> | </math> | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
u &= | u &=5-3x \\[2ex] | ||
du &= \frac{1}{ | du &=-3dx \\[2ex] | ||
-\frac{1}{3}du &=dx | |||
\end{align} | \end{align} | ||
| Line 17: | Line 17: | ||
\begin{align} | \begin{align} | ||
\int \frac{ | \int \frac{dx}{5-3x} &= -\frac{1}{3}\int \left(\frac{1}{u}\right)du \\[2ex] | ||
&=-\frac{1}{3}\ln{|u|} du \\[2ex] | |||
&= \ | &=-\frac{1}{3}\ln{|5-3x|}+C | ||
&= -\ | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 19:53, 22 September 2022
![{\displaystyle {\begin{aligned}u&=5-3x\\[2ex]du&=-3dx\\[2ex]-{\frac {1}{3}}du&=dx\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/084b592569fb6ffccbc984e9ef712892c68f0b7f)
![{\displaystyle {\begin{aligned}\int {\frac {dx}{5-3x}}&=-{\frac {1}{3}}\int \left({\frac {1}{u}}\right)du\\[2ex]&=-{\frac {1}{3}}\ln {|u|}du\\[2ex]&=-{\frac {1}{3}}\ln {|5-3x|}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/055a40ca0d1c3a8fd13120d0398782e1f212f655)