7.1 Integration By Parts/51b: Difference between revisions

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\begin{aligned}
\begin{aligned}
u &= \ln^{2}{x}  dv&=dx \\[2ex]
u & = \tfrac{1}{\sqrt{2}}(x+y) \qquad & x &= \tfrac{1}{\sqrt{2}}(u+v) \\[0.6ex]
du &= \frac{2}{x}\ln{(x)}dx  v&=x \\[2ex]
v & = \tfrac{1}{\sqrt{2}}(x-y) \qquad & y &= \tfrac{1}{\sqrt{2}}(u-v)
 
\end{aligned}
\end{aligned}
}
}

Revision as of 17:48, 29 November 2022