Perform partial fractions decomposition
1/((1+x)(3+x)) = A/(1+x)+B/(3+x)
1/((1+x)(3+x)) = (A(3+x)+B(1+x))/((1+x)(3+x))
1/((1+x)(3+x)) = ( (A+B)x +(3A+B) ) / ( (1+x) (3+x) )
{(A+B=0),(3A+B=1):}
{(A=-B),(-3B+B=1):}
{(A=1/2),(B=-1/2):}
So:
1/((1+x)(3+x)) = 1/(2(1+x))-1/(2(3+x))
Solve now:
dy/dx = y/((1+x)(3+x))
Separate the variables:
dy/y = dx/((1+x)(3+x))
lnabsy = int dx/((1+x)(3+x)) = 1/2 int dx/(1+x) -1/2int dx/(3+x)+C
lnabsy = 1/2 ln abs(1+x) -1/2ln abs(3+x) +C
For x > -1 both arguments are positive, so:
lnabsy = 1/2 ln (1+x) -1/2ln (3+x) = ln sqrt((1+x)/(3+x))+C
Taking the exponential:
absy = c_1 sqrt((1+x)/(3+x))
where c_1=e^C is positive. Admitting also negative value for c_1 and noting that y(x) = 0 is also a solution, then the general solution is:
y(x) = c sqrt((1+x)/(3+x)) with c in RR
We can determine the particular solution for:
y(1) = 2
c sqrt((1+1)/(3+1)) = 2
c sqrt(1/4) = 2
c/2= 2
c=4
Finally:
y(x) = 4sqrt((1+x)/(3+x))
