I want to ask about this problem. This problem is simple and I've already found the answer, but when I'm checked at wolfram, my solution is different with it.
Here's the problem:
Find the value of $$ \int \frac{1-\sqrt{3x+2}}{1+\sqrt{3x+2}}dx $$ ?
First of all, I need to make a substitution, in this case I use u as the substitutor.
Then, suppose $$ u= \sqrt{3x+2}$$
So, $$ u^2 = 3x+2 $$
$$ 2u\,\mathrm du = 3\,\mathrm dx $$
$$ \mathrm dx = \frac{2u\,\mathrm du}{3} $$
And the equation change to:
$$ \int \frac{(1-u)(2u)}{(1+u)(3)}\,\mathrm du $$
I simplify it to get:
$$ \frac{2}{3} \int \frac{u-u^2}{1+u}\,\mathrm du$$
Then I calculate it and get:
$$\frac23\left\{-\frac{u^2}2 +2u-2\ln|u+1|\right\}+C $$
And for the last, I substitute back the value of u to my answer to get:
$$-x-\frac23 +\frac{4\sqrt{3x+2}}{3}-\frac{4\ln|\sqrt{3x+2}+1|}{3}+ C $$
But, when I'm checking wolfram, the solution is not same. The constant is different.
I don't know where is my mistake. Can someone try to correct my mistake for this problem? Thanks
Let $y=\dfrac{1-\sqrt{3x+2}}{1+\sqrt{3x+2}}$
Now, let's try to find the inverse function $$\begin{align}y&=\dfrac{1-\sqrt{3x+2}}{1+\sqrt{3x+2}}\\\dfrac{y+1}{y-1}&=\dfrac{2}{-2\sqrt{3x+2}}\qquad\text{Componendo-Dividendo}\\\dfrac{1-y}{1+y}&=\sqrt{3x+2}\\3x+2&=\dfrac{(y-1)^2}{(y+1)^2}\\x&=\dfrac13\cdot\dfrac{(y-1)^2}{(y+1)^2}-\dfrac23\end{align}$$
Now, integrate by parts $$\begin{align}\int y\,\mathrm dx&=xy-\int x\,\mathrm dy\\&=xy-\int\dfrac13\cdot\dfrac{(y-1)^2}{(y+1)^2}\,\mathrm dy+\int\dfrac23\,\mathrm dy\\&=xy+\dfrac23y-\dfrac13\int\dfrac{(u-2)^2}{u^2}\,\mathrm du\qquad\qquad\qquad[\because u=y+1]\\&=xy+\dfrac23y-\dfrac13\int\dfrac{u^2-4u+4}{u^2}\mathrm du\\&=xy+\dfrac23y-\dfrac13\int\left(1-\dfrac4u+\dfrac4{u^2}\right)\,\mathrm du\\&=xy+\dfrac23y-\dfrac13y+4\ln|u|+\dfrac4u+C\\&=xy+\dfrac13y+4\ln|y+1|+\dfrac4{y+1}+C\end{align}$$
Substituting the value of $y$ ends the solution. Don't always worry about your answer to match with a CAS, you can do much better.