5.5 The Substitution Rule/13: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
No edit summary
Line 18: Line 18:


\int \frac{dx}{5-3x} &= -\frac{1}{3}\int \left(\frac{1}{u}\right)du \\[2ex]
\int \frac{dx}{5-3x} &= -\frac{1}{3}\int \left(\frac{1}{u}\right)du \\[2ex]
&=-\frac{1}{3}\int
&=-\frac{1}{3}\ln{e^3}


\end{align}
\end{align}
</math>
</math>

Revision as of 19:34, 22 September 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int \frac{dx}{5-3x} \text{,} \quad u=5-3x }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &=5-3x \\[2ex] du &=-3dx \\[2ex] -\frac{1}{3}du &=dx \end{align} }


Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int {\frac {dx}{5-3x}}&=-{\frac {1}{3}}\int \left({\frac {1}{u}}\right)du\\[2ex]&=-{\frac {1}{3}}\ln {e^{3}}\end{aligned}}}