I research this with Google, but apparently, it is so straightforward the websites give the answer as I already know that ...it is the arc Tan and on youtube one fellow substitutes the complex number for x and sure enough it works......

I can't help but think there is a straightforward way to do this with very minimal steps...and not using complex numbers. Does anyone out in the community know how?

  • 1
    $\begingroup$ Knowing the basic functions' derivatives is a big must to enter into antidifferentiation and integration. $\endgroup$
    – DonAntonio
    Jan 31, 2018 at 22:17
  • $\begingroup$ Maybe try a book ? $\endgroup$ Jan 31, 2018 at 22:17
  • 3
    $\begingroup$ Are you asking how to come up with the idea to substitute $x=\tan t$? $\endgroup$
    – J.G.
    Jan 31, 2018 at 22:18
  • $\begingroup$ You just have to be able to read a table of derivatives from right to left. Actually, this integral should be known by heart. $\endgroup$
    – Bernard
    Jan 31, 2018 at 22:27
  • $\begingroup$ @Bernard I think the OP is asking why this derivative is correct, and this is a question which can be answered. $\endgroup$ Jan 31, 2018 at 22:30

2 Answers 2


Here's how you prove that it's correct, if you've just plucked out of the air that the answer is $\tan^{-1}$ and you want to know why this is.

$$\int \frac{1}{1+x^2} \ \mathrm{d}x$$

Consider $\dfrac{\mathrm{d}}{\mathrm{d}x} \tan^{-1} \tan(x) = \sec^2(x) {\tan^{-1}}'(\tan(x))$ by the chain rule.

The left-hand side is just the derivative of $x$, so is $1$.

So multiplying both sides by $\cos(x)^2$, have $\cos(x)^2 = {\tan^{-1}}'(\tan(x))$.

So (writing $x = \tan^{-1}(u)$) have ${\tan^{-1}}'(u) = \cos(\tan^{-1}(u))^2$.

Now consider $\cos(x)^2 + \sin(x)^2 = 1$, so $1 + \tan(x)^2 = \sec(x)^2$; letting $x = \tan^{-1}(u)$, obtain $$1 + \tan(\tan^{-1}(u))^2 = \sec(\tan^{-1}(u))^2$$ where the left-hand side is just $1+u^2$, so $\cos(\tan^{-1}(u))^2 = \frac{1}{1+u^2}$.

Therefore ${\tan^{-1}}'(u) = \frac{1}{1+u^2}$, and hence integrating both sides $$\tan^{-1}(u) = \int \frac{1}{1+u^2} \ \mathrm{d}u$$

  • $\begingroup$ I will buy this ..thanx...I was hoping someone would know how to do it using a right triangle and making the hypotenuse x^2 + 1 and one of the sides 1..... $\endgroup$
    – Sedumjoy
    Jan 31, 2018 at 23:04

If you want more steps here's a step by step solution:

First make the substitution: $ x=\tan(\theta) $ | $dx = \sec^2(\theta)d\theta$

When plugging the new substituted variable you get:

$$\int \frac{\sec^2(\theta)}{\sec^2(\theta)}d\theta = \theta $$

Since $\theta=\arctan(x)$ the final answer is:

$$\int \frac{1}{x^2+1}dx = \arctan(x) + C $$

C is some arbitrary constant.


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