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 | &u = \ln(x)^{n} \quad dv= 1dx \\[2ex] | ||
du | &du =1dx \quad v=x \\[2ex] | ||
\end{align} | |||
</math> | |||
<math> | |||
\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} | \end{align} | ||
</math> | </math> |
Latest revision as of 04:36, 29 November 2022
Prove