Are $\frac{1}{6 \sqrt 2}\log \biggl|\frac{\sqrt 2 +1+3x}{\sqrt 2 -1-3x}\biggr|$ and $\frac{-1}{6 \sqrt 2}\log \biggl|\frac{ 3x+1-\sqrt 2}{3x +1 +\sqrt 2}\biggr|$ equal? How?
I am trying to integrate $ \int \frac{dx}{1-6x-9x^2}$ . I am getting the second expression as the answer. The book says it is the former one.
Yes $\frac{1}{6 \sqrt 2}\log \biggl|\frac{\sqrt 2 +1+3x}{\sqrt 2 -1-3x}\biggr|$ and $\frac{-1}{6 \sqrt 2}\log \biggl|\frac{ 3x+1-\sqrt 2}{3x +1 +\sqrt 2}\biggr|$ are equal. Proving this is just a matter of applying Logarithm properties. Using the fact $a\log b=\log b^a$ with $a=-1$ provides us with this equation $-\log(a)=\log\bigl(\frac{1}{a}\bigr)$.
$$\dfrac{-1}{6 \sqrt 2}\log \biggl|\frac{ 3x+1-\sqrt 2}{3x +1 +\sqrt 2}\biggr|=\dfrac{1}{6\sqrt{2}}\log\biggl|\dfrac{3x+1+\sqrt{2}}{3x+1-\sqrt{2}}\cdot\dfrac{(-1)}{(-1)}\biggr|$$Notice that doing so gives the desired denominator however to get the same denominator we must use the fact that $\log(a\cdot b)=\log(a)+\log(b)$.$$\dfrac{1}{6\sqrt{2}}\log\biggl|\dfrac{3x+1+\sqrt{2}}{-3x-1+\sqrt{2}}\cdot(-1)\biggr|=\dfrac{1}{6\sqrt{2}}\biggl[\log\bigl|-1\bigr|+\log\biggl|\dfrac{\sqrt{2}+1+3x}{\sqrt{2}-1-3x}\biggr|\biggr]$$Note that $\log\mid-1\mid=0$. Therefore the expression becomes $\frac{1}{6 \sqrt 2}\log \biggl|\frac{\sqrt 2 +1+3x}{\sqrt 2 -1-3x}\biggr|$. Hence they're equal.