7.1 Integration By Parts/51b: Difference between revisions
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du & = \tfrac{1}{x}dx & v &= x | du & = \tfrac{1}{x}dx & v &= x | ||
\end{aligned}} \\ [1ex] | \end{aligned}} \\ [1ex] | ||
&= x\ln(x)^3 -3x\ln^{2}{(x)} + 6\left[\ln{(x)}\cdot x - \int\dx\right] | |||
\end{align} | \end{align} | ||
</math> | </math> |
Revision as of 17:59, 29 November 2022
Failed to parse (unknown function "\dx"): {\displaystyle \begin{align} \int\ln(x)^3dx &= x\ln(x)^3 -\underbrace{3\int\ln(x)^2dx}_{ \begin{aligned} u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex] du & = \tfrac{2\ln{(x)}}{x}dx & v &= x \end{aligned} } \\ [1ex] &= x\ln(x)^3 -3\left[\ln^{2}{(x)}\cdot x - 2\int\ln{(x)}dx\right] \\ [1ex] &= x\ln(x)^3 -3x\ln^{2}{(x)} + \underbrace{6\int\ln{(x)}dx}_{ \begin{aligned} u & = \ln{(x)} & dv &= dx \\[0.6ex] du & = \tfrac{1}{x}dx & v &= x \end{aligned}} \\ [1ex] &= x\ln(x)^3 -3x\ln^{2}{(x)} + 6\left[\ln{(x)}\cdot x - \int\dx\right] \end{align} }