I'm cramming for a supplementary exam so you might see a ton of questions like these in the 48+ hours to come <3
The question is more of just a yes or no ; Evaluate the statement without calculating the indefinite integral.
$$ \int \frac{2x+1}{x+1} \,\mathrm dx = 2x -\ln|x+1| + C$$
1.) Seeing as how I'm not allowed to calculate the indefinite integral I assume I should work backwards using it. ie
$${{d} \over {dx}} (2x - \ln|x+1| + C)$$
2.) The hell am I supposed to do with $\ln|x+1|$, I have this tendency to split the page and answer the question due to $|x+1|$ having two possible answers.
That being said all I did was assume that $|x+1|$ would be $(x+1)$ and not $-(x+1)$. The result was :
$${2x+1} \over {x+1}$$
Does that answer the question? How would I go about showing my assumptions?
Oh and unrelated/related , I can do the above because of the fundamental theorem of calculus right?
You should know that $\dfrac d{dx}\ln|x|= \dfrac 1 x$.
To see that that is true, first do it piecewise: for $x>0$, and then for $x<0$. If $x<0$ then you have $\ln(-x)$, and $-x$ is positive, and you use the chain rule. Once you have done this just once, then remember it for use on subsequent occasions such as the exercise that you quote in your posted question.
Your technique is otherwise just what the person who posed the question probably expects.
There is one slight subtlety: The "constant" $C$ should be piecewise constant, i.e. one constant on the interval $x>-1$ and a possibly different constant on the interval $x<-1$, because there is a gap in the domain at $-1$.