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

Revision as of 04:30, 29 November 2022

Prove Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{}^{} \left(\ln(x)^{n}\right)dx = x\left(\ln(x)^{n}\right)-n\int_{}^{}\left(\ln(x)^{n-1}\right)dx }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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} }