\(\int \frac{x^2}{1+x^2}dx=\)1\(x+\tan^{-1}x+c\)2\(x^2+\tan^{-1}x+c\)3\(x-\tan^{-1}x+c\)4\(x^2-\tan^{-1}x+c\)Answer & explanation+Correct answer: option 3\(\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}\)
