The decomposition in partial fractions is
(3x^2+x+4)/((x^2+2)(x^2+1))=(Ax+B)/(x^2+2)+(Cx+D)/(x^2+1)3x2+x+4(x2+2)(x2+1)=Ax+Bx2+2+Cx+Dx2+1
3x^2+x+4=(Ax+B)(x^2+1)+(Cx+D)(x^2+2)3x2+x+4=(Ax+B)(x2+1)+(Cx+D)(x2+2)
Let x=0x=0 then 4= B+2D
Coefficients of x^2x2, 3=B+D3=B+D
From thes equations. we get D=1D=1 and B=2B=2
So,
3x^2+x+4=(Ax+2)(x^2+1)+(Cx+1)(x^2+2)3x2+x+4=(Ax+2)(x2+1)+(Cx+1)(x2+2)
We compare the coefficients of xx
1=A+2C1=A+2C
and coefficients of X^3X3, 0=A+C0=A+C
So, A=-1A=−1 and C=1C=1
(3x^2+x+4)/((x^2+2)(x^2+1))=(-x+2)/(x^2+2)+(x+1)/(x^2+1)3x2+x+4(x2+2)(x2+1)=−x+2x2+2+x+1x2+1
int((3x^2+x+4)dx)/((x^2+2)(x^2+1))=int((-x+2)dx)/(x^2+2)+int((x+1)dx)/(x^2+1)∫(3x2+x+4)dx(x2+2)(x2+1)=∫(−x+2)dxx2+2+∫(x+1)dxx2+1
=int(-xdx)/(x^2+2)+int(2dx)/(x^2+2)+int(xdx)/(x^2+1)+intdx/(x^2+1)=∫−xdxx2+2+∫2dxx2+2+∫xdxx2+1+∫dxx2+1
=-ln(x^2+2)/2+ln(x^2+1)/2+arctanx+sqrt2arctan(x/sqrt2)+C=−ln(x2+2)2+ln(x2+1)2+arctanx+√2arctan(x√2)+C
asintdx/(x^2+1)=arctanx∫dxx2+1=arctanx
and int(xdx)/(x^2+1)=ln(x^2+1)/2∫xdxx2+1=ln(x2+1)2
