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 | ||
Revision as of 06:58, 3 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}&\int {\frac {\left(\ln(x)\right)^{2}}{x}}dx\ =\ \int u^{2}du\\[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{aligned}}}