How do you integrate #(6x^2 + 1) /( (x^2)(x-1)^3)# using partial fractions?

1 Answer

#1/x+2/(x-1)-14/(x-1)^2+3\ln|(x-1)/x|+C#

Explanation:

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)#

#\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}#

#6x^2+1=(A+D)x^4+(-2A+B-3D+E)x^3+(A-B+C+3D-3E)x^2+(-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#