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Let $a>0$ and consider the integral \begin{equation} I(x) = \int_a^x \frac{1}{1+u} \, du. \end{equation}

On one hand, \begin{equation} I(x) = \ln(1+x) - \ln(1+a) = \ln\left(\frac{1+x}{1+a}\right). \end{equation}

And on the other, if we consider $u$ as $(\sqrt{u})^2 = |u| = u$ on $(a,x)$, \begin{equation} I(x) = \arctan(\sqrt{x}) - \arctan(\sqrt{a}). \end{equation}

This implies, \begin{equation} \ln\left(\frac{1+x}{1+a}\right) = \arctan(\sqrt{x}) - \arctan(\sqrt{a}). \end{equation}

Which seems like an interesting enough equality to feel like I should have seen something about it before. I mean, it does not seem at all obvious that natural log and inverse tangent should be related. Or is it? Am I missing something? Is there an error in my logic?

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    $\begingroup$ Have you tried testing that equality numerically with a few choices of $x$ and $a$? $\endgroup$ Dec 26, 2023 at 16:44
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    $\begingroup$ Even without evaluating numerically, the left-hand-side goes to infinity as $x$ goes to infinity, while the right-hand-side is bounded. $\endgroup$ Dec 27, 2023 at 5:42

5 Answers 5

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Your mistake is that in your arctan method you're asserting that $\left(f(\sqrt{x})\right)' = f'(\sqrt{x})$, since if you integrate both sides you get $$ f(\sqrt{x}) = \int \left(f(\sqrt{x})\right)' = \int f'(\sqrt{x}) $$ and taking $f(x) = \arctan(x)$ you get your second (erroneous) solution. This is incorrect, as the correct relation is $\left(f(\sqrt{x})\right)' = f'(\sqrt{x})\cdot \frac{1}{2\sqrt{x}}$ which is given by the chain rule. The way you undo the chain rule when integrating is doing a $u$-sub, which is (as others have pointed out) what you forgot to do.


Interestingly enough, there is a relationship between logarithms and the inverse tangent function, but it's a bit complex.

Defining the number $i = \sqrt{-1}$, then $i$ thus satisfies $i^2= -1$. Having this definition, by expanding the product as usual you can show that $x^2 +1 = (x+i)(x-i)$. Then, by partial fractions we get \begin{align*} \arctan(x)& = \int \frac{1}{x^2+1}\, \mathrm{d}x \\ &= \int \frac{1}{(x+i)(x-i)}\, \mathrm{d}x\\ &=\frac{i}{2} \int \frac{1}{x+i}\, \mathrm{d}x - \frac{i}{2} \int \frac{1}{x-i}\, \mathrm{d}x\\ & = \frac{i}{2}\ln(x+i) - \frac{i}{2}\ln(x-i) +C\\ & = \frac{i}{2}\ln\left(\frac{x+i}{x-i}\right)+C \end{align*} You can, in fact, find these kinds of logarithmic relations for all inverse trig functions, which at their core are possible because of Euler's formula $e^{ix} = \cos(x) +i \sin(x)$ which relates trig functions with exponentials, thus also relating their inverses.

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    $\begingroup$ Remark: The expression $\arctan(x)$ is analgous to $\text{arctanh}(x)=\frac 12\log(\frac{1+x}{1-x})$. $\endgroup$
    – Divide1918
    Dec 26, 2023 at 5:45
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    $\begingroup$ A bit complex, is it? $\endgroup$
    – No Name
    Dec 26, 2023 at 17:24
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If you differentiate $f(x) = \arctan(\sqrt x)$ you will find that $f'(x)$ is not $\frac{1}{1+x}$ (because of the chain rule). So the second method is incorrect.

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Your second method makes the substitution $v=\sqrt u$, so you should get $du=2v dv\ne dv$, hence the substitution would not yield an integrand of $\dfrac{1}{1+v^2}$.

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$$I(x) = \int_a^x \frac{1}{1+u} \, du$$

Cross checking by differentiation of OP's second result: $$\frac{d}{du}(\arctan\sqrt u){=\frac{1}{1+(\sqrt u)^2}\cdot\frac{d}{du}(\sqrt u)\quad (\text{By chain rule})\\=\frac{2\sqrt u}{1+u}}\\ \therefore \int \frac{1}{1+u} \, du\ne \arctan \sqrt u +c \,\\ \text{And, } \int \frac{2\sqrt u}{1+u} \, du= \arctan \sqrt u +c \,$$

Explaining the correct way second process:

$$\sqrt{u}=:y\iff \frac{1}{2\sqrt{u}}du=dy\iff du=2ydy\\\int \frac{1}{1+u} du{=\int \frac{2ydy}{1+y^2}\\=\int \frac{d(1+y^2)}{1+y^2}\\=\ln|1+y^2|+c\\=\ln|1+u|+c}$$ Therefore, $$I(x) = \int_a^x \frac{1}{1+u}=\ln(1+x) - \ln(1+a) = \ln\left|\frac{1+x}{1+a}\right|$$

OP's mistake: Changing of variable.

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The answer with the logarithm is correct.

For the answer you obtain with the arctangent, you want to write $u = (\sqrt u)^2$ so you may want to perform a change of variables $t = \sqrt u$. Redo your working and see if the answer you obtained is correct or not.

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