0

I was integrating the function $f(x)=\frac{x^2}{x-1}$ with two methods: 1) by starting with $u$-substitution and 2) by starting with polynomial long division.

Upon my attempt at method 1, I let $u=x-1$ and after evaluating and subbing back the $x-1$ term back in, I got the result of $\frac{x^2}{2}+x-\frac{3}{2}+\ln |x-1|+C$.

With method two I did polynomial long division to obtain that $\frac{x^2}{x-1}=x+1+\frac{1}{x-1}$. Integrating this I obtained the result of $\frac{x^2}{2}+x+\ln |x-1|+C$.

I suppose that these must be the same answers as they are both are valid methods, so in method 1, can the $-\frac{3}{2}$ term just be amalgamated into the $+C$ as it is a constant?

Physics Fusion
  • 57
  • 1
    They are different constant. The first constant will be $\frac{3}{2}$ more than the second one – Lorago Mar 16 '25 at 18:16
  • so is "method 1" incorrect? – Physics Fusion Mar 16 '25 at 18:17
  • No, it is correct. You can have as many constants as possible in your integral, so long as you remember that when differentiated, the constants will equal $0$. In your case, you could call $-3/2 + C$ the constant $K$ and when you differentiate your function, you should get the same function. I'll elaborate with a longer comment. – bjcolby15 Mar 16 '25 at 18:22
  • Both answers are correct. However. they will have different values for C. Consider your first answer. If you let $C_1 = C - 3/2$ and then substitute in $C_1$ for $C - 3/2$ you should realize that you have the same answer. – Bob Mar 16 '25 at 18:22
  • Thank you for the prompt help. So if I were to do a definite integral from here, I would just plug in the bounds into $\frac{x^2}{2}+x-\frac{3}{2}+\ln |x-1|$, so no $\frac{-3}{2}$. Is my understanding correct? – Physics Fusion Mar 16 '25 at 18:26
  • Suggestion: if you're going to be doing more than one integration, write $C_1$,$C_2$, etc., from the very beginning. – JonathanZ Mar 16 '25 at 18:31

1 Answers1

2

Yes, both integrations are correct, and you can combine the constants $-\dfrac {3}{2} + C$ into another constant (say $K$), as long as you get back the original function when differentiating as the derivative of a constant is $0.$

In your case, when we differentiate $$\frac{x^2}{2}+x-\frac{3}{2}+\ln |x-1|+C$$ we get $$x+1-0+\dfrac {1}{x-1} + 0$$ or $$(x+1) + \dfrac {1}{x-1} = \dfrac {(x+1)(x-1)+1}{x-1} = \dfrac {x^2}{x-1}.$$ When we differentiate $$\frac{x^2}{2}+x+\ln |x-1|+C$$ we get $$x+1+\dfrac {1}{x-1}$$ which is also $\dfrac {x^2}{x-1}.$

bjcolby15
  • 4,210