# The solution of the differential equation $\large\frac{dy}{dx}+\frac{2xy}{(1+x^2)}=\frac{1}{(1+x^2)^2}$ is

$(A)\;y(1+x^2)=c+\tan^{-1}x \quad (B)\;\frac{y}{1+x^2}=c+\tan^{-1}x \quad(C)\;ylog(1+x^2)=c+\tan^{-1}x \quad (D)\;y(1+x^2)=c+\sin^{-1}x$

Toolbox:
• If a differential equation of the form: $\large\frac{dy}{dx}$$+Py=Q, then its general solution is given by ye^{\int pdx}=\int Q.e^{\int pdx}+c Given \large\frac{dy}{dx}+\frac{2xy}{(1+x^2)}=\frac{1}{(1+x^2)^2} This is of the form:\large\frac{dy}{dx}$$+Py=Q$
Where $P=\large\frac{2x}{1+x^2}$$\;and \; Q=\large\frac{1}{(1+x^2)^2} \int pdx=\int \large\frac{2x}{1+x^2}$$dx$
Put $(1+x^2)=t$ On differentiating w.r.t x we get, $2xdx=dt$
Hence substituting this we get,
$\large\frac{dt}{t}$$=\log t=\log |1+x^2| Hence the integraing factor e^{\int pdx}=e^{\log (1+x^2)} =(1+x^2) Hence the required solution is y \times (I.F)=\int Q.(I.F)dx+c y(1+x^2)=\int =\large\frac{1}{(1+x^2)^2}$$\times (1+x^2)dx+c$
$y(1+x^2)=\int =\large\frac{1}{1+x^2} $$dx+c \int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$ \tan ^{-1} \bigg(\large\frac{x}{a}\bigg)$
On integrating we get,
$y(1+x^2)=\tan^{-1}(x)+c$
Hence the correct option is $A$