How to prove that : $\displaystyle \int \frac{1}{1+x^2} \, \mathrm{d}x = \arctan(x) + C$?

Question

The question is the title... It would be nice that the demonstration doesn't start from the result but from the indefinite intergral

Answer 1

Well it depends on what you do want to use (or can use), but if it isn't a standard integral you can rely on; substitute $x = \tan t$, then use $1+\tan^2t = \sec^2t$ and $x'(t) = \sec^2t$, so: $$\int \frac{1}{1+x^2} \, \mbox{d}x \to \int \frac{1}{1+\tan^2t} x'(t)\, \mbox{d}t = \int \frac{1}{\sec^2t} \sec^2t\, \mbox{d}t= \int \, \mbox{d}t = t + C$$ Now if $x = \tan t$, then $t = \ldots$.

Answer 2

I have always taught this by first finding the derivative of $\arctan x$. Or is that what you where trying to avoid?

Let $y=\arctan x$ so $\tan y = x$. Then implicitly differentiate to get:

$$\sec^2 y \cdot y' = 1$$

$$y'=\cos^2y$$

Then consider a right angle triangle with angle $y$ in one corner. As $\tan y=x$ then the adjacent side can be $1$ and the opposite side can be $x$. Using Pythagoras gives the hypotenuses as $\sqrt{1+x^2}$. So $\cos y=\frac{1}{\sqrt{1+x^2}}$ and hence:

$$y'=\frac{1}{1+x^2}$$

The integral then comes from the fundamental theorem of calculus.