Josuem95981@students.laalliance.org: Created page with " $\begin{align} &\int_4^9 \frac{\ln(y)}{y}dy \ \ \ \ \ \ u= \ln(y), du=\frac{1}{y}dy \ \ \ \ dv=\frac{1}{\sqrt y}dy, v=2\sqrt y \\[2ex] \\ & u\cdot v \ - \int v du \ = \ \big(\ln(y) \big) \cdot \big( 2\sqrt y \big) - \int \big(2\sqrt y \ \cdot \frac{1}{y} \big)dy \\[2ex] &= 2\sqrt y \ln(y) - \int\frac{2}{\sqrt y }dy \ = \ 2\sqrt y \ln(y) - 2\int \frac{1}{\sqrt y}dy \ = \ 2\sqrt y \ln(y) - 2\int \frac{1}{\sqrt y} \\[2ex] &=2\sqrt y \ln(y)-2 \cdot 2\sqrt y \ = \..."$

2022-11-29T05:56:52Z

Created page with "<math> \begin{align} &\int_4^9 \frac{\ln(y)}{y}dy \ \ \ \ \ \ u= \ln(y), du=\frac{1}{y}dy \ \ \ \ dv=\frac{1}{\sqrt y}dy, v=2\sqrt y \\[2ex] \\ & u\cdot v \ - \int v du \ = \ \big(\ln(y) \big) \cdot \big( 2\sqrt y \big) - \int \big(2\sqrt y \ \cdot \frac{1}{y} \big)dy \\[2ex] &= 2\sqrt y \ln(y) - \int\frac{2}{\sqrt y }dy \ = \ 2\sqrt y \ln(y) - 2\int \frac{1}{\sqrt y}dy \ = \ 2\sqrt y \ln(y) - 2\int \frac{1}{\sqrt y} \\[2ex] &=2\sqrt y \ln(y)-2 \cdot 2\sqrt y \ = \..."

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<math>
\begin{align}

&\int_4^9 \frac{\ln(y)}{y}dy \ \ \ \ \ \ u= \ln(y), du=\frac{1}{y}dy \ \ \ \ dv=\frac{1}{\sqrt y}dy, v=2\sqrt y \\[2ex]
\\

& u\cdot v \ - \int v du \ = \ \big(\ln(y) \big) \cdot \big( 2\sqrt y \big) - \int \big(2\sqrt y \ \cdot \frac{1}{y} \big)dy \\[2ex]
&= 2\sqrt y \ln(y) - \int\frac{2}{\sqrt y }dy \ = \ 2\sqrt y \ln(y) - 2\int \frac{1}{\sqrt y}dy \ = \ 2\sqrt y \ln(y) - 2\int \frac{1}{\sqrt y} \\[2ex]
&=2\sqrt y \ln(y)-2 \cdot 2\sqrt y \ = \ 2\sqrt y \ln(y)-4\sqrt y \\[2ex]
&= 2\sqrt y \big(\ln(y) -2) \bigg|_4^9 \\[2ex]
\\

&=\bigg( 2 \sqrt9 \big(\ln(9)-2 \big) \bigg) - \bigg(2\sqrt4 \big(\ln(4)-2\big) \bigg) \ = \ \bigg(6\ln(9)-12 \bigg)-\bigg(4\ln(4)-8\bigg) \\[2ex]
&=\ln\big(9^6\big)-4 + \ln\big(4^{-4}\big) \ = \ \ln\big(9^6 \cdot 4^{-4}\big)-4 \ = \ \ln\bigg(\frac{9^6}{4^4}\bigg)-4 \ = \ 2\ln\bigg(\frac{729}{16}\bigg) -4 \\[2ex]
&=4\ln\bigg(\frac{27}{4}\bigg)-4

\end{align}
</math>

7.1 Integration By Parts/22 - Revision history