MCQ
સમીકરણ $(1 + {x^2})\frac{{dy}}{{dx}} = 1$ નો ઉકેલ મેળવો.
- A$y = \log (1 + {x^2}) + c$
- B$y + \log (1 + {x^2}) + c = 0$
- C$y - \log (1 + x) = c$
- ✓$y = {\tan ^{ - 1}}x + c$
==>$\frac{{dy}}{{dx}} = \frac{1}{{1 + {x^2}}}$
On integrating, $y = {\tan ^{ - 1}}x + c$.
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