Juanr69976@students.laalliance.org at 23:03, 26 November 2022

2022-11-26T23:03:48Z

← Older revision		Revision as of 23:03, 26 November 2022
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	<math> f'(x)= \int_{1}^{2} x^4(\ln(x))^2 \cdot dx </math> <br><br>		<math> f'(x)= \int_{1}^{2} x^4(\ln(x))^2 \cdot dx </math> <br><br>

	<math>\int_{1}^{2} x^4(\ln(x))^2 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2}{5}\int_{1}^{2} x^4 \ln(x)\cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(\ln(x))^2}{25} + \int_{1}^{2} x^4 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(ln(x))^2}{25} + \frac{2x^5}{25} </math><br>		<math>\int_{1}^{2} x^4(\ln(x))^2 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2}{5}\int_{1}^{2} x^4 \ln(x)\cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(\ln(x))^2}{25} + \int_{1}^{2} x^4 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(ln(x))^2}{25} + \frac{2x^5}{25} \Bigg\|_1^2</math><br>
	<math> u= (\ln(x))^2 ~ ~ ~ ~ ~ dv=dx ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ u=\ln(x) ~ ~ ~ ~ ~ dv=x^4</math> <br>		<math> u= (\ln(x))^2 ~ ~ ~ ~ ~ dv=dx ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ u=\ln(x) ~ ~ ~ ~ ~ dv=x^4</math><br>
	<math> du= \frac{2}{x} \ln(x) \cdot dx ~ ~ v=x ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ du=\frac{1}{x} ~ ~ ~ ~ ~ v=\frac{x^5}{5}</math> <br><br>		<math>du= \frac{2}{x} \ln(x) \cdot dx ~ ~ v=x ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ du=\frac{1}{x} ~ ~ ~ ~ ~ v=\frac{x^5}{5}</math> <br><br>
	<math>=\frac{32}{5} (\ln(2))^2 -\frac{64}{25} (\ln(2)) + \frac{62}{125}</math>		<math>=\frac{32}{5} (\ln(2))^2 -\frac{64}{25} (\ln(2)) + \frac{62}{125}</math>

Juanr69976@students.laalliance.org: Created page with " $f'(x)= \int_{1}^{2} x^4(\ln(x))^2 \cdot dx$

$\int_{1}^{2} x^4(\ln(x))^2 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2}{5}\int_{1}^{2} x^4 \ln(x)\cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(\ln(x))^2}{25} + \int_{1}^{2} x^4 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(ln(x))^2}{25} + \frac{2x^5}{25}$
$u= (\ln(x))^2 ~ ~ ~ ~ ~ dv=dx ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ u=\ln(x) ~ ~ ~ ~ ~ dv=x^4$
2022-11-26T23:00:03Z

Created page with "<math> f'(x)= \int_{1}^{2} x^4(\ln(x))^2 \cdot dx </math> <math>\int_{1}^{2} x^4(\ln(x))^2 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2}{5}\int_{1}^{2} x^4 \ln(x)\cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(\ln(x))^2}{25} + \int_{1}^{2} x^4 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(ln(x))^2}{25} + \frac{2x^5}{25} </math> <math> u= (\ln(x))^2 ~ ~ ~ ~ ~ dv=dx ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ u=\ln(x) ~ ~ ~ ~ ~ dv=x^4</m..."

New page
<math> f'(x)= \int_{1}^{2} x^4(\ln(x))^2 \cdot dx </math> 

<math>\int_{1}^{2} x^4(\ln(x))^2 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2}{5}\int_{1}^{2} x^4 \ln(x)\cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(\ln(x))^2}{25} + \int_{1}^{2} x^4 \cdot dx ~ ~ ~ = ~ ~ ~ \frac{x^5(ln(x))^2}{5} - \frac{2x^5(ln(x))^2}{25} + \frac{2x^5}{25} </math> 
<math> u= (\ln(x))^2 ~ ~ ~ ~ ~ dv=dx ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ u=\ln(x) ~ ~ ~ ~ ~ dv=x^4</math> 
<math> du= \frac{2}{x} \ln(x) \cdot dx ~ ~ v=x ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ du=\frac{1}{x} ~ ~ ~ ~ ~ v=\frac{x^5}{5}</math> 
<math>=\frac{32}{5} (\ln(2))^2 -\frac{64}{25} (\ln(2)) + \frac{62}{125}</math>

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