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] | ||
& u=\ln(x) \\ | |||
& 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
![{\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/14c67c7e2833b4725e660286233fe1fe30b4f5db)