7.1 Integration By Parts/51b: Difference between revisions

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<math> \text{Use exercise 47 to evaluate} \int(\ln{x})^3dx </math> <br><br>
<math> \text{Use } \int(\ln{(x)}^{n} = x(\ln{x})^n - n\int(\ln{x})^{n-1}dx </math> <br><br><br>
<math> \text{Exercise 47: } x(\ln{x})^n-n\int(\ln{x})^{n-1}dx </math> <br><br>


<math>
<math>
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\begin{align}
\begin{align}


\int\ln(x)^3dx &= x\ln(x)^3 -3\underbrace{\int\ln(x)^2dx}_{
\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}} \\ [2ex]
 
&= 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}_{
\begin{aligned}
\begin{aligned}
u &= \ln^{2}{x} \qquad dv&=dx \\[2ex]
u & = \ln{(x)} & dv &= dx \\[0.6ex]
du &= \frac{2}{x}\ln{(x)}dx \quad v&=x \\[2ex]
du & = \tfrac{1}{x}dx & v &= x
\end{aligned}} \\ [2ex]


\end{aligned}
&= 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


\end{align}
\end{align}


</math>
</math>

Latest revision as of 18:21, 29 November 2022