The value of _ -1 ^ 1 1+ x 1+x^ 2 d x is :
Mathematics · JEE Main · NTA Exams — Integral Calculus
The value of \(\int_{-1}^{1} \frac{1+\sin x}{1+x^{2}} d x\) is :
- \(\frac{\pi}{4}\)
- \(-\frac{\pi}{4}\)
- \(\frac{\pi}{2}\)
- \(-\frac{\pi}{2}\)
Answer
(C) 2
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