Need help with Differential equation concept Having trouble with this differential equation.
Find the general solution to the diﬀerential equation:
$$ \frac {dy}{dx}= \frac {√y(x^2 + x − 4)} {(x^2 + 1)(x − 1)} $$
I don't know where to start.
 A: $$\text{y}'\left(x\right)=\frac{\sqrt{\text{y}(x)}(x^2+x-4)}{(x^2+1)(x-1)}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x=\int\frac{x^2+x-4}{(x^2+1)(x-1)}\space\text{d}x\tag1$$
Substitute $\text{u}=\text{y}\left(x\right)$ and $\text{d}\text{u}=\text{y}'\left(x\right)\space\text{d}x$ to find the integral $\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x$:
$$\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x=2\sqrt{\text{y}\left(x\right)}+\text{C}_1\tag2$$
For the RHS use partial fractions:
$$\frac{x^2+x-4}{(x^2+1)(x-1)}=\frac{2x}{x^2+1}+\frac{3}{x^2+1}-\frac{1}{x-1}\tag3$$
So, you get that:
$$\int\frac{x^2+x-4}{(x^2+1)(x-1)}\space\text{d}x=\ln\left|x^2+1\right|+3\arctan(x)-\ln\left|x-1\right|+\text{C}_2\tag4$$
So, we know that:
$$2\sqrt{\text{y}\left(x\right)}=\ln\left|x^2+1\right|+3\arctan(x)-\ln\left|x-1\right|+\text{C}\tag5$$
Solving for $\text{y}\left(x\right)$:
$$\text{y}\left(x\right)=\left\{\frac{\ln\left|\frac{x^2+1}{x-1}\right|+3\arctan(x)}{2}+\text{C}\right\}^2\tag6$$
