WolframAlpha gives a real closed form for this definite integral:
∫+∞−∞arctan(11+x2)dx=√2(√2−1)π
Yet, the formula it gives for the indefinite integral uses i.
∫tan−1(1x2+1)dx=xtan−1(1x2+1)+2(tan−1(x√1−i)(1−i)3/2+tan−1(x√1+i)(1+i)3/2)+C
Why isn't the definite integral non-real?
Answer
First of all, it is real because it converges and because arctan(11+x2) is a real number for every x∈R.
On the other hand, that some number can be represented with an expression involving i doesn't mean it is not a real number. For instance,
1i+i=0∈R.
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