5.5 The Substitution Rule/19: Difference between revisions
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& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex] | & = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex] | ||
u=\ln(x) | & u=\ln(x) \\ | ||
du=\frac{1}{x}dx | & du=\frac{1}{x}dx \\ | ||
& = \ \frac{1}{3}(\ln(x))^3+C | & = \ \frac{1}{3}(\ln(x))^3+C | ||
Latest revision as of 06:59, 3 September 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} & \int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du \\[2ex] & = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex] & u=\ln(x) \\ & du=\frac{1}{x}dx \\ & = \ \frac{1}{3}(\ln(x))^3+C \end{align} }