7.1 Integration By Parts/37: Difference between revisions
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\end{align} | \end{align} | ||
</math> <br> | </math> <br> | ||
<math>\int(u-1)\ln({u}) du </math> <br><br> | |||
<math> w = \ln{u} \qquad dv = (u-1)dv </math> <br><br> | |||
<math> du = \frac{1}{u}du \qquad v = \frac{1}{2}u^2-u </math> <br><br> | |||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
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&= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 + \frac{1}{2}x + \frac{3}{4} + c \\[2ex] | &= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 + \frac{1}{2}x + \frac{3}{4} + c \\[2ex] | ||
\end{align} | \end{align} | ||
< | </math> | ||
Latest revision as of 03:09, 29 November 2022
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} u &= x + 1 \\[2ex] x &= u-1 \\[2ex] du &= dx \\[2ex] \end{align} }
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(u-1)\ln({u}) du }
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 w = \ln{u} \qquad dv = (u-1)dv }
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 du = \frac{1}{u}du \qquad v = \frac{1}{2}u^2-u }
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(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u - \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex] &= \ln{u} \cdot \frac{1}{2}u^2 - u - \int(\frac{1}{2}u - 1)du \\[2ex] &= \ln{u} \cdot \frac{1}{2}u^2 - u - [\frac{1}{4} u^2 - u] + c \\[2ex] &=\ln({1+x})(\frac{1}{2}(x^2+2x+1)-(1+x)) - (\frac{1}{4}(1+x)^2-(1+x)) + c \\[2ex] &= \ln({1+x})(\frac{1}{2}x^2 + x + \frac{1}{2} - x- 1) - (\frac{1}{4}(1+x)^2-(1+x)) + c \\[2ex] &= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 - \frac{1}{2}x-\frac{1}{4}+1+x+c \\[2ex] &= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 + \frac{1}{2}x + \frac{3}{4} + c \\[2ex] \end{align} }
