Why is the antiderivative of $\frac{1}{1+x^2}=\tan^{-1}(x)$? My textbook says the antiderivative of $\frac{1}{1+x^2}$ is $\tan^{-1}(x)$. 
To confirm this to myself I took the derivative of $\tan^{-1}(x)$ expecting to get $\frac{1}{1+x^2}$ , but instead I ended up with $-\frac{1}{\sin^2(x)}$. So why is $\tan^{-1}(x)$ the antiderivative of $\frac{1}{1+x^2}$ if the  derivative of $\tan^{-1}(x)$ is not $\frac{1}{1+x^2}$? Shouldn't the derivative of the antiderivative of a function give you the original function?
 A: $$
\begin{align}
\arctan(x) &= y\\
x &= \tan(y)\\
\frac{\mathrm d}{\mathrm dx} x &= \frac{\mathrm d}{\mathrm dx} \tan(y)\\
1 &= y' \sec^2(y)\\
y'&=\dfrac{1}{\sec^2(y)}\\
y'&=\dfrac{1}{\tan^2(y)+1}\\
y'&=\dfrac{1}{x^2 + 1}
\end{align}
$$
Because $x = \tan(y)$
A: $$x=\arctan y$$
$$y=\tan x$$
$$\frac {dy}{dx}=\sec^2x=1+\tan^2 x=1+y^2$$
$$\frac {dx}{dy}=\frac {1}{1+y^2}$$
A: $$
y = \tan x
$$
$$
\frac {dy}{dx} = \sec^2 x = 1+\tan^2 x = 1+y^2
$$
$$
\frac{dy}{dx} = 1+y^2
$$
$$
\frac{dx}{dy} = \frac{1}{1+y^2}
$$
$$
\frac d{dx} \arctan y = \frac 1 {1+y^2}
$$
Is the reciprocal of $dy/dx$ really $dx/dy$?  It certainly would be if $dy$ and $dx$ were actual numbers.  The fact that these are each others reciprocal's is an instance of the chain rule.
A: Consider the following.
$$ x=\tan(y)$$
$$dx = \sec^2(y)dy$$
Therefore.
$$\int\frac{1}{1+x^2}dx=\int\frac{\sec^2(y)}{1+\tan^2(y)}dy$$
$$=\int\frac{\sec^2(y)}{\sec^2(y)}dy$$
$$=\int dy$$
$$=y+c$$
$$=\arctan(x)+c$$
A: Using the formula for geometric series,
$${1\over 1+x^2}=1-x^2+x^4-x^6+\cdots+(-x^2)^n+\cdots
$$
Integrating,
$$
\begin{align}
\int 1-x^2+x^4-x^6+\cdots+(-x^2)^n+\cdots dx&=x-{x^3\over3}+{x^5\over5}-{x^7\over7}+\cdots\\
&=\arctan x\\
\end{align}
$$
A: Generally speaking,$f^{-1}$ is the inverse of $f(x)$, not $\frac{1}{f(x)}$.
For example, $\tan^{-1}x=\arctan x$, $\sin^{-1}x=\arcsin x$.
