Given:
∫4x1/x^{2} +3x+2 dx
Let 4x1 = A.d/dx(x^{2} +3x+2) + B
4x1=A(2x+3) + B
Equating the coefficients of x and constant terms on both sides:
4x= 2xA 3A+B=1
A=2 B=7
∫4x1/x^{2} +3x+2 dx = ∫2(2x+3)  7 dx/x^{2} +3x+2
=2∫2x3 dx/x^{2} +3x+2 7∫dx/x^{2} +3x+2
=2I_{1}  7I_{2}
For I_{1} :
Let x^{2} +3x+2 = z
2x+3 dx = dz
I_{1} : 2∫dz/z = 2logz = 2logx^{2} +3x+2 +C
For I_{2} :
7∫dx/x^{2} +3x+2 = 7∫dx/(x+3/2)^{2}  (1/2)^{2}
7/1/2*2 logx+3/21/2/x+3/2+1/2 +C'
7logx+1/x+2 +C'
I_{1} +I_{2} = 2logx^{2} +2x+3 7logx+1/x+2+A where A=C+C' (Constant of integration)
indraniadhikari12345 2 hours ago
