7.1 Integration By Parts/54: Difference between revisions
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du &= \frac{1}{x} dx \quad v=x \\ | du &= \frac{1}{x} dx \quad v=x \\ | ||
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\int_{1}^{5} \left(x\ln(x) \right)dx = \frac{x^2\ln(x)}{2}\bigg|_{1}^{5} - \int_{1}^{5} \left(\frac{x^2}{2x} \right)dx = \frac{x^2\ln(x)}{2}\bigg|_{1}^{5} - \frac{1}{2}\int_{1}^{5} \left(x \right)dx\\ | \int_{1}^{5} \left(x\ln(x) \right)dx = \frac{x^2\ln(x)}{2}\bigg|_{1}^{5} - \int_{1}^{5} \left(\frac{x^2}{2x} \right)dx = \frac{x^2\ln(x)}{2}\bigg|_{1}^{5} - \frac{1}{2}\int_{1}^{5} \left(x \right)dx\\ | ||
Revision as of 04:01, 29 November 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} u &= \ln(x) \quad dv= 1 dx \\ du &= \frac{1}{x} dx \quad v=x \\ } Failed to parse (syntax error): {\displaystyle \int_{1}^{5} \left(x\ln(x) \right)dx = \frac{x^2\ln(x)}{2}\bigg|_{1}^{5} - \int_{1}^{5} \left(\frac{x^2}{2x} \right)dx = \frac{x^2\ln(x)}{2}\bigg|_{1}^{5} - \frac{1}{2}\int_{1}^{5} \left(x \right)dx\\ u &= \ln(x) \quad dv= x dx \\ du &= \frac{1}{x} \quad v=\frac{x^2}{2} \\ \end{align} }