<s>[INST] Solution of differential equation $x^3dy -x\ln x dx = (1+x^2) \tan^{-1}x dx -x dy $.... Solution of differential equation 
$x^3dy -x \ln x dx = (1+x^2) \tan^{-1}x dx -x dy $ ( options are ) 
(a) $y = \frac{\ln x}{\tan^{-1}x}+c$ 
(b)  $y = \tan^{-1}x \ln x +c$ 
(c)  $y =\frac{\tan^{-1}x}{\ln x }+c$ 
(d)  none of these 
Solution : The given equation can be written as : 
$$\frac{dy}{dx} =\frac{(1+x^2)\tan^{-1}x +x \ln x }{x^3+x} \Rightarrow \int dy = \int ( \frac{\tan^{-1}x}{x} +\frac{\ln x}{1+x^2})dx $$ 
Let $$ \frac{\tan^{-1}x}{x} = I \Rightarrow I = \tan^{-1} \int ( \frac{1}{x})dx + \int ( d (\tan^{-1}x) \int \frac{1}{x}dx$$ .... Please suggest how to integrate this function as it is again coming the same integral. Thanks... 
 [/INST] It follows from integration by parts that $$\int\frac{\tan^{-1}x}{x}dx =\tan^{-1}x\ln x-\int \frac{\ln x}{1+x^2}dx$$ and hence we have $$\int\frac{\tan^{-1}x}{x}dx+\int \frac{\ln x}{1+x^2}dx=\tan^{-1}x\ln x+c.$$ Therefore, $y=\tan^{-1}x\ln x+c$ is the solution.
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