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?
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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.
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 $$\tan^{-1}(u) = \int \frac{1}{1+u^2} \ \mathrm{d}u$$