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- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,678

- Mar 1, 2012

- 655

$u = 1+ \ln(y^2) = 1+2\ln{y} \implies du = \dfrac{2}{y} \, dy$

$\displaystyle \dfrac{1}{2} \int \dfrac{du}{u} = \dfrac{1}{2}\ln|u|+ C$

$\displaystyle \dfrac{1}{2} \ln|1+\ln(y^2)| + C$

- Thread starter
- #3

- Jan 31, 2012

- 2,678

ok I dont think I see what happened between the first and second step

$$dy=\frac{y}{2}du$$

never mind I see what it is

$$dy=\frac{y}{2}du$$

never mind I see what it is

Last edited:

- Mar 1, 2012

- 655

- Jan 30, 2018

- 426

$u= 1+ ln(y^2)= 1+ 2ln(y)$

$u- 1= 2 ln(y)$

$\frac{u-1}{2}= ln(y)$

$e^{\frac{u-1}{2}}= y$.