Is $\int\frac{1}{x^2+1}\,dx$ an even or odd function? Is $$\int\frac{1}{x^2+1}\,dx$$ an even or odd function?
Because $\int\frac{1}{(x-1)^2+1}\,dx=\int\frac{1}{(1-x)^2+1}\,dx$, it appears to be even.
Also because $\int\frac{1}{x^2+1}\,dx=\arctan(x)$, it appears to be odd.
 A: $$ \int\frac{1}{x^2+1}\,dx=\arctan(x)+c $$
is not a function, but a collection of functions.
The element of the collection corresponding to $c=0$ is an odd function. The others are neither even nor odd.
A: The inside function is even.
The integral is an odd function in $t$ if you consider it to be the integral from 0 to $t$. Without specifying the starting point of the integral, though, it is not necessarily either even or odd.
Notice that the integral is in fact $\arctan(x)+C$ if you don't specify the bounds.
A: In general, the indefinite integral of an even function is odd (plus some constant).
Let $f(x) = f(-x)$, with an antiderivative $F(x)$. Break the antiderivative $F(x)$ into even part $E(x)$ and odd part $O(x)$:
$$\begin{cases}F(x) = E(x) + O(x)\\
F(-x) = E(-x) + O(-x) = E(x) - O(x)
\end{cases}
$$
So the even part is 
$$\begin{align*}
E(x) &= \frac{F(x) + F(-x)}2\\
&= F(0) + \frac{F(x) -  F(0) + F(-x) - F(0)}2\\
&= F(0) + \frac12\left[\int_0^xf(u)\ du + \int_0^{-x}f(u)\ du\right]\\
&= F(0) + \frac12\left[\int_0^xf(u)\ du - \int_0^x f(-v)\ dv\right]&&(u=-v)\\
&= F(0) + \frac12\left[\int_0^x f(u)\ du - \int_0^xf(v)\ dv\right]\\
&= F(0)
\end{align*}
$$
The odd part is
$$\begin{align*}
O(x) &= F(x) - E(x)\\
&= F(x) - F(0)\\
&= \int_0^x f(u)\ du
\end{align*}$$
i.e. a particular antiderivative is an odd function, when


*

*adjusted by a constant so that $F(0) = 0$, or when

*obtained by definite integral $F(x) = \int_0^x f(u)\ du$.

A: As for another part of my confusion, my classmate gave me this:
$$\int\frac{1}{(1-x)^2+1}\,d(-x)=-\int\frac{1}{(x-1)^2+1}\,dx$$
without constant $C$, it is odd.
Thank all four of you, my several days of confusion dissipated.
