7.1 Integration By Parts/11: Difference between revisions

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</math>
</math>


<math>
\begin{align}
\begin{align}
u=1+16t^{2} \\[1ex]
u=1+16t^{2} \\[1ex]

Revision as of 05:10, 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}in(u)= \text {tarctan(4t)}-\frac{1}{8}in(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} }