7.1 Integration By Parts/47: Difference between revisions

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Prove
<math>
<math>
\int_{}^{} \left(\ln(x)^{n}\right)dx = x\left(\ln(x)^{n}\right)-n\int_{}^{}\left(\ln(x)^{n-1}\right)dx
\int_{}^{} \left(\ln(x)^{n}\right)dx = x\left(\ln(x)^{n}\right)-n\int_{}^{}\left(\ln(x)^{n-1}\right)dx


</math>
<math>
\int_{}^{} \left(\ln(x)^{n}\right)dx
</math>
</math>


<math>
<math>
\begin{align}
\begin{align}
u &= \ln(x)^{n} \quad dv= 1dx \\[2ex]
&u = \ln(x)^{n} \quad dv= 1dx \\[2ex]
du &=1dx        \qquad v=x \\[2ex]
&du =1dx        \quad v=x \\[2ex]


\end{align}
\end{align}
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<math>
<math>
\int_{}^{} \left(\ln(x)^{n}\right)dx = x \ln(x)^{n} - \int_{}^{} \left((\frac{n \ln(x)^{n-1}}{x}) \rigt)dx
\begin{align}


\int_{}^{} \left(\ln(x)^{n}\right)dx &= x \ln(x)^{n} - \int_{}^{} \left((x \frac{n \ln(x)^{n-1}}{x}) \right)dx \\[2ex]
&= x \ln(x)^{n} - \int_{}^{} \left(n \ln(x)^{n-1} \right)dx \\[2ex]
&= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex]
\end{align}
</math>
</math>

Latest revision as of 04:36, 29 November 2022

Prove