Since the function is not defined at 0, the definition set is not a connex. Thus, the set of primitives is not the result of this computation. To get the general form of a primitive, you have to allow the C to change from one connex composant to the other (one C for x1). That's why I would have given a definite integral instead.
When I did it myself I got 2ln|-2x+2|-0.5x+c, and I thought that I was wrong but it's all good, because it's a constant. Demonstration: 2ln|-2x+2|-0.5x+c = 2 * (ln|(-2)*(x-1)|)-0.5x+c = 2 * (ln(-2)+ln|x-1|)) - 0.5x + c = 2ln|x-1| - 0.5 x + (c + 2ln(-2)) = 2ln|x-1| - 0.5x + c. The last c is a different one, of course. It still blows my mind.
Since the function is not defined at 0, the definition set is not a connex. Thus, the set of primitives is not the result of this computation. To get the general form of a primitive, you have to allow the C to change from one connex composant to the other (one C for x1). That's why I would have given a definite integral instead.
When I did it myself I got 2ln|-2x+2|-0.5x+c, and I thought that I was wrong but it's all good, because it's a constant. Demonstration:
2ln|-2x+2|-0.5x+c = 2 * (ln|(-2)*(x-1)|)-0.5x+c =
2 * (ln(-2)+ln|x-1|)) - 0.5x + c =
2ln|x-1| - 0.5 x + (c + 2ln(-2)) =
2ln|x-1| - 0.5x + c.
The last c is a different one, of course.
It still blows my mind.
I had the same problem! 😀
bruh i forgot how to do long divsion lol
Sir please solve integral of x^3/(x+1)^2by long division method