Let
\frac{6x^2+1}{x^2(x-1)^3}=A/(x-1)+B/(x-1)^2+C/(x-1)^3+D/x+{E}/(x^2)6x2+1x2(x−1)3=Ax−1+B(x−1)2+C(x−1)3+Dx+Ex2
\frac{6x^2+1}{x^2(x-1)^3}=frac{Ax^2(x-1)^2+Bx^2(x-1)+Cx^2+(Dx+E)(x-1)^3}{x^2(x-1)^3}6x2+1x2(x−1)3=Ax2(x−1)2+Bx2(x−1)+Cx2+(Dx+E)(x−1)3x2(x−1)3
6x^2+1=(A+D)x^4+(-2A+B-3D+E)x^3+(A-B+C+3D-3E)x^2+(-D+3E)x-E6x2+1=(A+D)x4+(−2A+B−3D+E)x3+(A−B+C+3D−3E)x2+(−D+3E)x−E
Comparing the corresponding coefficients on both the sides we get
A+D=0\ .........(1)
-2A+B-3D+E=0\ .........(2)
A-B+C+3D-3E=6\ ............(3)
-D+3E=0\ ............(4)
E=-1\ ..........(5)
Solving all above five linear equations, we get
A=3, B=-2, C=7, D=-3, E=-1
Now, setting above values , the partial fractions are given as follows
\frac{6x^2+1}{x^2(x-1)^3}=3/(x-1)-2/(x-1)^2+7/(x-1)^3-3/x-1/x^2
Now, integrating above equation w.r.t. x, we get
\int \frac{6x^2+1}{x^2(x-1)^3}\ dx=\int 3/(x-1)\ dx-\int2/(x-1)^2\ dx+\int 7/(x-1)^3\ dx-\int 3/x\ dx-\int 1/x^2\ dx
=3\int \frac{dx}{x-1}-2\int (x-1)^{-2}\ dx+7\int (x-1)^{-3}\ dx-3\int dx/x-\int x^{-2}\ dx
=3\ln|x-1|+2/(x-1)-14/(x-1)^2-3\ln|x|+1/x+C
=1/x+2/(x-1)-14/(x-1)^2+3\ln|(x-1)/x|+C
