Find $\int{\arctan x}\,\mathrm dx$ without substitution

Question

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

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

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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

• Integration of arctan(x) is itself?
• Indefinite integral of $\arctan(x)$, why consider $1\cdot dx$?

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

Answer 1

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

Answer 2

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

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

Answer 3

$\newcommand{\dx}{\mathrm dx\,}$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 $\log(1+x)$ $$\log(1+x)=\sum\limits_{n\geq1}(-1)^{n-1}\frac {x^n}n$$