The solution of \(\frac{1}{x} \cdot \frac{d y}{d x}=\tan ^{-1} x\) is……
A. \(\frac{x^{2} \tan ^{-1} x}{2}+c=0\)
B. x tan-1x + c = 0
C. x – tan-1x = c
D. \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)
Answer:D
The solution of \(\frac{1}{x} \cdot \frac{d y}{d x}=\tan ^{-1} x\) is……
A. \(\frac{x^{2} \tan ^{-1} x}{2}+c=0\) B. x tan-1x + c = 0 C. x – tan-1x = c D. \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)