7.1 Integration By Parts/51b: Difference between revisions
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\int\ln(x)^3dx &= x\ln(x)^3 -\underbrace{3\int\ln(x)^2dx}_{ | \int\ln(x)^3dx &= x\ln(x)^3 -\underbrace{3\int\ln(x)^2dx}_{ | ||
\begin{aligned} | \begin{aligned} | ||
u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex] | u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex] | ||
du & = \tfrac{2\ln{(x)}}{x}dx & v &= x | du & = \tfrac{2\ln{(x)}}{x}dx & v &= x | ||
\end{aligned} | \end{aligned}} \\ [2ex] | ||
} \\ [ | |||
&= x\ln^{3}(x) -3\left[\ln^{2}{(x)}\cdot x - 2\int\ln{(x)}dx\right] \\ [1ex] | &= x\ln^{3}(x) -3\left[\ln^{2}{(x)}\cdot x - 2\int\ln{(x)}dx\right] \\ [1ex] | ||
&= x\ln^{3}(x) -3x\ln^{2}{(x)} + \underbrace{6\int\ln{(x)}dx}_{ | &= x\ln^{3}(x) -3x\ln^{2}{(x)} + \underbrace{6\int\ln{(x)}dx}_{ | ||
\begin{aligned} | \begin{aligned} | ||
u & = \ln{(x)} & dv &= dx \\[0.6ex] | u & = \ln{(x)} & dv &= dx \\[0.6ex] | ||
du & = \tfrac{1}{x}dx & v &= x | du & = \tfrac{1}{x}dx & v &= x | ||
\end{aligned}} \\ [ | \end{aligned}} \\ [2ex] | ||
&= x\ln^{3}(x) -3x\ln^{2}{(x)} + 6\left[\ln{(x)}\cdot x - \int dx\right] \\[1ex] | &= x\ln^{3}(x) -3x\ln^{2}{(x)} + 6\left[\ln{(x)}\cdot x - \int dx\right] \\[1ex] | ||
&= x\ln^{3}(x) -3x\ln^{2}{(x)} + 6x\ln{(x)} - 6x + C | &= x\ln^{3}(x) -3x\ln^{2}{(x)} + 6x\ln{(x)} - 6x + C | ||
\end{align} | \end{align} | ||
</math> | </math> |
Latest revision as of 18:21, 29 November 2022