MHB Solve for y: APC.7.3.01 Int Subs

[karush]

ok hope the image is readable
kinda odd with the y is in the denominator
but was going to start with $u=1+\ln y^2 \quad y=?? $
not sure ?

 

[skeeter]

no need to solve for $y$ ...

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

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

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

 

[karush]

ok I don't think I see what happened between the first and second step

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

never mind I see what it is

 

[skeeter]

\[ \int \dfrac{dy}{y[1+\ln(y^2)]} = \dfrac{1}{2} \int \dfrac{1}{{\color{blue}1+2\ln{y}}} \cdot {\color{red}\dfrac{2}{y} \, dy} = \dfrac{1}{2} \int \dfrac{1}{{\color{blue}u}} \, {\color{red}du} \]

 

[HOI]

Yes, you don't need to solve for y. But if you did it is just algebra:
$u= 1+ ln(y^2)= 1+ 2ln(y)$
$u- 1= 2 ln(y)$
$\frac{u-1}{2}= ln(y)$
$e^{\frac{u-1}{2}}= y$.