5.5 The Substitution Rule/19: Difference between revisions
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(Created page with "<math> \begin{align} \int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C ---- = \ \frac{1}{3}(\ln(x))^3+C \end{align} </math>") |
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\begin{align} | \begin{align} | ||
\int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du | & \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 | & = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex] | ||
---- | ---- | ||
= \ \frac{1}{3}(\ln(x))^3+C | & = \ \frac{1}{3}(\ln(x))^3+C | ||
Revision as of 06:53, 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]----&=\ {\frac {1}{3}}(\ln(x))^{3}+C\end{aligned}}}