Find the function y solution of $$(x+1)(x^2+1)y'=2x^2+x$$
Initial conditions : $y = 1$ when $x=0 $ I tried to simplify the equation first by breaking down the fraction in two sides. so that I will find the primitive of both side then find y. But my problem is the second part $(3x-1)/(2x^2+2)$. I cannot find its primitive.
On
We have $y'=\frac{2x^2+x}{(x+1)(x^2+1)}$. Assuming $\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$ for some $A,B,C \in \mathbb{R}$ and equating the coefficients of the respective powers of $x$, we'll get $A=\frac{1}{2}, B=\frac{3}{2}, C=-\frac{1}{2}$.
Therefore $y'=\frac{1}{2(x+1)}+\frac{3x}{2(x^2+1)}-\frac{1}{3(x^2+1)}$, which gives integrating both side, $$y=\frac{1}{2}\ln (x+1)+\frac{3}{4}\ln (x^2+1)-\frac{1}{2}\tan^{-1}x+a,\quad \text{ where } a \text { is an arbitrary constant.}$$
Now, as $y=1$ for $x=0$, so $a=1$ and thus $$y=\frac{1}{2}\ln (x+1)+\frac{3}{4}\ln (x^2+1)-\frac{1}{2}\tan^{-1}x+1.$$
As Adrian Kleister mentioned within the comments and which was in fact your own attempt is to seperate the equation in the following way
$$\begin{align} (x+1)(x^2+1)y′=2x^2+x\Leftrightarrow y′=\frac{2x^2+x}{(x+1)(x^2+1)} \end{align}$$
Taking the integral leads to
$$y=\int \frac{2x^2+x}{(x+1)(x^2+1)} dx +c$$
where the integral can be solved by using the technique of partial decomposition. Therefore we write
$$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}\Leftrightarrow2x^2+x=A(x^2+1)+(Bx+C)(x+1)$$
From hereon we get a system of equations for $A,B$ and $C$. Solving this system yields to $A=\frac12$, $B=\frac32$ and $C=-\frac12$ and so we have to integrate now
$$y=\int \frac{3x-1}{2(x^2+1)}+\frac{1}{2(x+1)} dx +c = \frac12\int \frac{3x}{x^2+1}-\frac{1}{x^2+1}+\frac{1}{x+1} dx +c$$
The solution to this can be obtained by using standard techniques based on the natural logarithm and the inverse tangens. So finally the anti-derivatives and so the function $y$ is given by
$$y=\frac12 \ln(x+1) + \frac34\ln(x^2+1) - \frac12\arctan(x) + c $$
Plugging in the IVP yields to
$$1=c$$
and therefore the solution is given by
$$y(x)=\frac12 \ln(x+1) + \frac34\ln(x^2+1) - \frac12\arctan(x) + 1 $$