MCQ
$\int \frac{1}{\left(x^2+1\right)^2} d x=$
- A$\tan ^{-1} x-\frac{1}{2 x\left(x^2+1\right)}+ c$
- B$\tan ^{-1} x+\frac{1}{2\left(x^2+1\right)}+ c$
- ✓$\frac{1}{2} \tan ^{-1} x+\frac{x}{2\left(x^2+1\right)}+ c$
- D$\tan ^{-1} x+\frac{1}{x^2+1}+ c$
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