MCQ
વિકલ સમીકરણ $(1 + {x^2})\frac{{dy}}{{dx}} = x$ નો ઉકેલ મેળવો.
- A$y = {\tan ^{ - 1}}x + c$
- B$y = - {\tan ^{ - 1}}x + c$
- ✓$y = \frac{1}{2}{\log _e}(1 + {x^2}) + c$
- D$y = - \frac{1}{2}{\log _e}(1 + {x^2}) + c$
==> $\int {dy} = \int {\frac{x}{{1 + {x^2}}}dx} + c$ ==> $y = \frac{1}{2}{\log _e}(1 + {x^2}) + c$.
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