7.1 Integration By Parts/11: Difference between revisions

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\begin{align}
\begin{align}


& u=1+16t^{2} \\[1ex]
& \qquad u=1+16t^{2} \\[1ex]
& du=32t dt \\[1ex]
& du=32t dt \\[1ex]
& \frac{1}{32}du=t dt  
& \frac{1}{32}du=t dt  

Revision as of 05:13, 29 November 2022