Trying to study for my exams, and stumpled across a Integration that i could not solve myself. I looked up the teachers solution, and I dont get how he reduced the integral.
The image is of an integral that has been simplified. I want to know the steps to perform this reduction, calculations rules etc.
On
The specific step in the linked image follows from two things.
First, we have that $1=\frac{2}{2}=\frac{2}{-1}\cdot(-\frac{1}{2})$
Second, we have $-x\sqrt{4\ln(2x-1)}=-x\cdot\sqrt{4}\cdot\sqrt{\ln(2x-1)}=-2x\sqrt{\ln(2x-1)}$
The second further implies that $\sqrt{\ln(2x-1)}-x\sqrt{4\ln(2x-1)}=\color{blue}{\sqrt{\ln(2x-1)}}-2x\color{blue}{\sqrt{\ln(2x-1)}}=1\cdot \color{blue}{\sqrt{\ln(2x-1)}} - 2x\cdot \color{blue}{\sqrt{\ln(2x-1)}}$
$ = (1-2x)\cdot\color{blue}{\sqrt{\ln(2x-1)}}$
We have as a result $\frac{1}{\sqrt{\ln(2x-1)}-x\sqrt{4\ln(2x-1)}}=\frac{1}{(1-2x)\sqrt{\ln(2x-1)}}$
"Multiplying by one" using the first observation we have:
$\frac{1}{(1-2x)\sqrt{\ln(2x-1)}}=\frac{1}{(1-2x)\sqrt{\ln(2x-1)}}\cdot 1=\frac{1}{(1-2x)\sqrt{\ln(2x-1)}}\cdot \frac{2}{-1}\cdot (-\frac{1}{2})$
$=(-\frac{1}{2})\cdot \frac{2}{-1\cdot(1-2x)\sqrt{\ln(2x-1)}}=(-\frac{1}{2})\cdot\frac{2}{(2x-1)\sqrt{\ln(2x-1)}}$
We have then finally:
$$\int\limits_1^{\frac{e+1}{2}}\left(\frac{1}{\sqrt{\ln(2x-1)}-x\sqrt{4\ln(2x-1)}}\right)dx = \int\limits_1^{\frac{e+1}{2}}\left((-\frac{1}{2})\cdot\frac{2}{(2x-1)\sqrt{\ln(2x-1)}}\right)dx$$
as the integrands are equal. You may then move the $(-\frac{1}{2})$ outside of the integral as integrals are linear operators.
HINT:
Factoring the term $\sqrt{\log(2x-1)}$ yields
$$\frac{1}{\sqrt{\log(2x-1)}-x\sqrt{4\log(2x-1)}}=\frac{1}{\sqrt{\log(2x-1)}\,(1-2x)}$$