MCQ
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$x \tan ^{-1} x+c=0$
- C$x-\tan ^{-1} x=c$
- ✓$y=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c$
$\operatorname{Hint}: \frac{1}{x} \frac{d y}{d x}=\tan ^{-1} x \quad \therefore d y=x \tan ^{-1} x d x$
$\therefore \int d y=\int\left(\tan ^{-1} x\right) \cdot x d x$
$\therefore y=\left(\tan ^{-1} x\right) \cdot \frac{x^2}{2}-\int \frac{d}{d x}\left(\tan ^{-1} x\right) \cdot \frac{x^2}{2} d x+c$
$=\frac{x^2 \tan ^{-1} x}{2}-\int \frac{1}{1+x^2} \times \frac{1}{2} x^2 d x+c$
$=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{\left(1+x^2\right)-1}{1+x^2} d x+c$
$=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int\left(1-\frac{1}{1+x^2}\right) d x+c$
$\left.=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c.\right]$
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