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If you compute the integral:
$$ \int \frac{2x}{(x+1)^2}.$$ by substitution (using $u=x+1$, then you will get.

$$ 2 \ln | 1 + x | + \frac{2}{1+x} + C.$$

But if, instead, you use integration by parts: $u=x, dv=2/(1+x)^2$, you will get

$$ -\frac{2x}{1+x} + 2 \ln | 1 +x | + C .$$

Clearly $C$ cannot depend on $x$. The two results should be the same.

Why are they different?

Herman Jaramillo
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  • The answers are in fact equivalent. – Adam Rubinson Dec 21 '23 at 17:05
  • @AdamRubinson : I do not see that. Could you please explain? – Herman Jaramillo Dec 21 '23 at 17:07
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    $$\frac{2}{1+x}=2-\frac{2x}{1+x}$$ – sreysus Dec 21 '23 at 17:09
  • This is a good illustration of why I think indefinite integrals should be banned. Especially the formula for the indefinite integral of $1/x$. You never see indefinite integrals in any setting after Calculus 2, not even in Calculus 3. You do use an antiderivative when using the Fundamental Theorem of Calculus, but you need only one specific one, no $+C$ needed. It's the definite integral that matters, and all of the attention should be focused on that. Then there's no need to figure out what exactly "+C$ means. – Deane Dec 21 '23 at 17:18
  • @Deane The name "indefinite integral" is lousy: that must be called, imo, antiderivative. And we use it a lot after Cals. 2: to evaluate definite (Riemann) integrals, say. – DonAntonio Dec 21 '23 at 17:48
  • @DonAntonio, an indefinite integral is not an antiderivative. It is a misguided attempt to represent all antiderivatives. I did mention the use of an antiderivative for the Fundamental Theorem of Calculus. – Deane Dec 21 '23 at 17:57
  • @Deane It is exactly THE ANTIDERIVATIVE (or (all) the antiderivatives if you will)...which is this weird being, not a function at all, that represents each and every function that is an antiderivative of the integrand function. – DonAntonio Dec 21 '23 at 18:12

3 Answers3

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The difference of the two functions you get (with different $\;C\;$ each) is:

$$2\log|1+x|+\frac2{1+x}+C-\left(-\frac{2x}{1+x}+2\log|1+x|+C'\right)=$$

$$=\frac{2+2x}{1+x}+\overbrace{K}^{=C-C'}=2+K=\text{ a constant}$$

and thus both of them are a primitive function of $\;\cfrac{2x}{(1+x)^2}\;$ .

DonAntonio
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The solution is not a function. It is a collection of functions. You can think of it as a list of functions. Each different value of $C$ is a solution (a member of the list). The two solutions look different, but they give the same collection of functions. If you write $2\ln|1+x|+\frac{2}{1+x}+C_1$ and $\frac{-2x}{1+x}+2\ln|1+x|+C_2$, you can set these equal and solve. That is, for each $C_1$ in the first function, you'll get a $C_2$ in the second function, and vice versa. But the constants won't be the same.

If you think about $C$ as one number, it looks confusing. But if you remember that $C$ ranges over all values, you'll see that you get the same list.

  • In your second expression, you have $-2/(1+x)$, while the OP is saying it is $-2x/(1+x)$. –  Dec 21 '23 at 17:07
  • @jwhite : First, one is $2/(1+x)$ and the other $-2x/(1+x).$ If you dump those into C1 and C2, you would get something like $C_2 = C_1 + 2/(1+x) + 2x/(1+x)$ and $C_2$ depends on $x$. That cannot happen. – Herman Jaramillo Dec 21 '23 at 17:11
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    @HermanJaramillo $\frac{2}{1+x}+\frac{2x}{1+x}=\frac{2(1+x)}{1+x}=2$ doesn't depend on $x$. –  Dec 21 '23 at 17:30
  • @Doug Thanks for spotting that. I've corrected it. –  Dec 21 '23 at 17:31
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    @jwhite : you are correct. I failed to see that. Thanks. – Herman Jaramillo Dec 21 '23 at 21:05
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$$ -\frac{2x}{1+x} = \frac{-2x}{1+x} = \frac{-2x-2+2}{1+x} = \frac{-2(1+x)}{1+x} + \frac{2}{1+x} = -2 + \frac{2}{1+x}.$$

Adam Rubinson
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