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]
} \\ [1ex]


&= 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}} \\ [1ex]
\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