2018-08-15: I'm still looking for an answer that does not rely on

$$\int{f\left[g(x)\right]g'(x)}\,\mathrm dx = F\left[g(x)\right]$$

I'm refreshing my old calculus skills, and the textbook (Kalkulus by Tom Lindstrøm, 3rd edition, a Norwegian book) asks me to find $\int{\arctan x}\,\mathrm dx$.

I start with integration by parts:

$$\int{\arctan x}\,\mathrm dx = \int{1\cdot\arctan x}\,\mathrm dx = x\cdot\arctan x - \int{x\cdot \frac{1}{x^2+1}}\,\mathrm dx$$

Next, it would be natural to use substitution, and I can do that to get the answer:

$$x\cdot\arctan x - \frac{\ln(x^2+1)}{2}+c$$

Which is correct, but it's clear from context that the book wants me to do it without using substituation. I feel like I've tried everything, looking at the book's own examples, but I must be missing some essential trick.

Here are some of the approaches I've tried:

1. Integration by parts, again

1.1. $u = x$ , $v' = \frac{1}{x^2+1}$

$$x\cdot\arctan x - \left(x\cdot\arctan x - \int{1\cdot\arctan x}\,\mathrm dx\right) = \int{\arctan x}\,\mathrm dx$$

Back where I started.

1.2. $u = \frac{1}{x^2+1}$ , $v' = x$

$$x\cdot\arctan x - \left(\frac{x^2}{2} \cdot \frac{1}{x^2+1} - \int{\frac{x^2}{2} \cdot \frac{-2x}{(x^2+1)^2}}\,\mathrm dx\right)$$ $$= x\cdot\arctan x - \frac{x^2}{2(x^2+1)} - \int{\frac{x^3}{(x^2+1)^2}}\,\mathrm dx$$

I've tried hacking away at this, but it doesn't look like it's getting any easier.

2. Adding and subtracting $x^2$ in the numerator

$$x\cdot\arctan x - \int{x\cdot \frac{(x^2+1)-x^2}{x^2+1}}\,\mathrm dx$$ $$= x\cdot\arctan x - \int{x\cdot \left(1 - \frac{x^2}{x^2+1}\right)}\,\mathrm dx$$ $$= x\cdot\arctan x - \int{x - x\cdot\frac{x^2}{x^2+1}}\,\mathrm dx$$ $$= x\cdot\arctan x - \frac{x^2}{2} + \int{x\cdot\frac{x^2}{x^2+1}}\,\mathrm dx$$

The book uses a similar trick (adding and subtracting 1 in the numerator) to solve $\int{x\cdot\arctan x}\,\mathrm dx$. In the process it finds that $\int{\frac{x^2}{x^2+1}}\,\mathrm dx = x - \arctan x$. Perhaps I need to use this result:

Partial integration with $u = x$ , $v´ = \frac{x^2}{x^2+1}$

$$= x\cdot\arctan x - \frac{x^2}{2} + \left(x\cdot(x-\arctan x) - \int{1\cdot (x-\arctan x)}\,\mathrm dx\right)$$ $$= x\cdot\arctan x - \frac{x^2}{2} + \left(x^2 - x\cdot\arctan x - \frac{x^2}{2} + \int{\arctan x}\,\mathrm dx\right)$$ $$= x\cdot\arctan x - \frac{x^2}{2} + x^2 - x\cdot\arctan x - \frac{x^2}{2} + \int{\arctan x}\,\mathrm dx$$ $$= \int{\arctan x}\,\mathrm dx$$

Back where I started, again. I've been trying for days, different tricks and manipulations. The book doesn't even list it as a particularly tricky question, so I'm feeling a bit dumb.

3. Searching for the answer

I've looked at some other questions, including

But they don't seem to answer my particular question.

  • $\begingroup$ @Zacky No expert, but yes $\endgroup$
    – S.T. Veje
    Aug 15, 2018 at 20:45

3 Answers 3


I don't know if this is what you are looking for or not, but functions $f$ and their inverses $f^{-1}$ are pretty cool. If you know the derivative of $f$, you know the derivative of $f^{-1}$. And if you know the antiderivative of $f$, you know the antiderivative of $f^{-1}$ as well via this interesting formula on integration of inverse functions . I write it out here

$$ \int f^{-1}(y) \;dy = yf^{-1}(y) - F \;\circ \; f^{-1}(y) + C$$

So if you know that the antiderivative of $\tan{x}$ is $-\ln(\cos{x})$, then the result follows. This is a trick, so I'm not sure it's what you are looking for. Also, in order to find the antiderivative of $\tan{x}$, you use u-substitution which goes against our goal of integrating without u-substitution. Anyways, hopes this helps

  • 1
    $\begingroup$ This is cool. I can't say for certain if it's what the book wanted me to do, probably not, but I like this solution. I also think I'm allowed to just "know" the antiderivative of $\tan x$ in this case. $\endgroup$
    – S.T. Veje
    Aug 16, 2018 at 15:03

$\newcommand{\dx}{\mathrm dx\,}$The only other way I can see without using a u-substitution, which must I mention, is the easiest way to evaluate this integral, is using the infinite geometric sequence. Hopefully you remember that$$\sum\limits_{n\geq0}(-1)^n\, x^{2n}=\frac 1{1+x^2}$$ Therefore, calling the integral $\mathfrak{I}$, then $$\begin{align*}\mathfrak{I} & =\int\dx\frac x{1+x^2}\\ & =\sum\limits_{n\geq0}(-1)^n\int\dx x^{2n+1}\\ & =\frac 12\sum\limits_{n\geq1}(-1)^{n-1}\frac {x^{2n}}n\\ & =\frac 12\log(1+x^2)+C\end{align*}$$ Where in the second to last line, we’ve made use of the taylor expansion series for $\log(1+x)$ $$\log(1+x)=\sum\limits_{n\geq1}(-1)^{n-1}\frac {x^n}n$$

  • $\begingroup$ If I may, but why does your book require the general rule, no substitutions? It honestly seems counter intuitive and overcomplicates the problem... $\endgroup$
    – Frank W
    Aug 15, 2018 at 20:18
  • $\begingroup$ I'm starting to suspect this exercise slipped into an early section of the book by mistake and was meant to appear later. It's so obviously supposed to be solved with substitution, I've just been holding out hope that there was some simple trick I'd missed. I don't like to assume the book made a mistake $\endgroup$
    – S.T. Veje
    Aug 15, 2018 at 20:34
  • $\begingroup$ A "no substitution" rule doesn't make any sense, really, because (i) "I just spotted $x/(x^2+1)$ is the derivative of $\frac{1}{2}\ln(x^2+1)$" is equivalent to using substitution and (ii) you could argue any inference of an antiderivative boils down to "substituting $u=$ the answer gets $\int du$". $\endgroup$
    – J.G.
    Aug 15, 2018 at 20:52
  • $\begingroup$ @J.G I’m still having trouble understanding your second point of substituting $u$ as the answer. Can you elaborate that? $\endgroup$
    – Frank W
    Aug 15, 2018 at 22:03
  • $\begingroup$ @FrankW Substituting $u=x\arctan x-\tfrac{1}{2}\ln(1+x^2)$ gives $\int\arctan x dx=\int du = u + C = x\arctan x-\tfrac{1}{2}\ln(1+x^2) + C$. If I just write down the second equation, I must have realised I could have deduced it from the first. $\endgroup$
    – J.G.
    Aug 16, 2018 at 5:45

Observe that $$I=\int{x\cdot \frac{1}{x^2+1}}\,\mathrm dx$$ can be rewritten as

$$I=\frac{1}{2}\int{\frac{\mathrm d(x^2+1)}{x^2+1}}=\ln(x^2+1)$$

  • $\begingroup$ It's $\frac{x}{x^2+1}$ (plus, not minus). Can you factor $x^2+1$? However, the book hasn't discussed partial fractions yet (same as substitution), so I don't think I'm meant to use that either. $\endgroup$
    – S.T. Veje
    May 28, 2018 at 12:27
  • $\begingroup$ I don't know why you want to factorize that since $d(x^2+1)=2x$. So,basically just multiply Nr. and Dr. by 2 and you have straightforward integral. This problem can be done in the same way. Nothing of substitution; just straight forward integration. $\endgroup$
    – Your IDE
    May 28, 2018 at 12:31
  • $\begingroup$ I'm not following (though I'll keep trying). Note that my question had a small mistake in the first integral, it should have been a plus not a minus in the denominator (fixed now). $\endgroup$
    – S.T. Veje
    May 28, 2018 at 12:40
  • $\begingroup$ @SebastianS.T.Veje I changed it. $\endgroup$
    – Your IDE
    May 28, 2018 at 12:45
  • 1
    $\begingroup$ The problem is found in the section discussing partial integration, which is before the section discussing substitution (including that formula), so it would be very odd if it expects you to use substitution (or that formula) to solve it. There must be some clever trick to do it without. $\endgroup$
    – S.T. Veje
    May 28, 2018 at 12:57

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