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 (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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} }