7.1 Integration By Parts/11: Difference between revisions
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
& u=1+16t^{2} | & u=1+16t^{2} \\ [1ex] | ||
& du=32t dt | & du=32t dt \\ [1ex] | ||
& \frac{1}{32}du=t dt | & \frac{1}{32}du=t dt | ||
\end{align} | \end{align} | ||
</math | </math> | ||
Revision as of 05:27, 29 November 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 \text {arctan(4t)}dt }
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 u= \text {arctan(4t)} \qquad dv=dt }
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 du= \frac{4}{1+(4t)^{2}} dt \qquad v=t }
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 \text {arctan(4t)}dt = \text {tarctan(4t)}-4\int \frac{t}{1+16t^{2}} dt = \text {tarctan(4t)}-\frac{4}{32}\int\frac{1}{u} = \text {tarctan(4t)}-\frac{1}{8}\ln(u)= \text {tarctan(4t)}-\frac{1}{8}\ln(1+16t^{2})+C }
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=1+16t^{2} \\ [1ex] & du=32t dt \\ [1ex] & \frac{1}{32}du=t dt \end{align} }