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

Question

Emilia Aguilar · Accepted Answer

The integral is #int((6x^2+1)dx)/(x^2(x-1)^2)=-1/x+2lnx-2ln(x-1)-7/(x-1) +C#

First, let's figure out the partial fractions' coefficients.

#(6x^2+1)/(x^2(x-1)^2)=A/x^2+B/x+C/(x-1)+D/(x-1)^2#

#(6x^2+1)/(x^2(x-1)^2)=(A(x-1)^2+Bx(x-1)^2+Cx^2(x-1)+Dx^2)/(x^2(x-1)^2)#

#(6x^2+1)=A(x-1)^2+Bx(x-1)^2+Cx^2(x-1)+Dx^2#

Let x=0, then #1=A+0+0+0# ; So #A=1# Let x=1, then #7=D# ; so #D=7# Also,coefficients of #x^3#; #0=B+C#, ; so #B=-C# And coefficients of #x^2# ; #6=A-2B-C+D# So #6=1+2C-C+7# #=>##C=-2# And #B=2#

#(6x^2+1)/(x^2(x-1)^2)=1/x^2+2/x-2/(x-1)+7/(x-1)^2#

So

#int((6x^2+1)dx)/(x^2(x-1)^2)=int(1dx)/x^2+int(2dx)/x-int(2dx)/(x-1)+ (7dx)/(x-1)^2#

#=-1/x+2lnx-2ln(x-1)-7/(x-1) +C#

Scarlett Allison · Answer

undefined To integrate (6x^2 + 1)/(x^2(x - 1)^2) using partial fractions, first, express the rational function as a sum of partial fractions.

The partial fraction decomposition of (6x^2 + 1)/(x^2(x - 1)^2) is:

(6x^2 + 1)/(x^2(x - 1)^2) = A/x + B/x^2 + C/(x - 1) + D/(x - 1)^2

To find the values of A, B, C, and D, you can multiply both sides by the common denominator x^2(x - 1)^2:

6x^2 + 1 = A(x - 1)^2 + Bx(x - 1)^2 + Cx^2(x - 1) + Dx^2

After expanding and simplifying, equate coefficients of like terms to find A, B, C, and D.

Once you have found the values of A, B, C, and D, rewrite the original expression as a sum of the partial fractions. Then, integrate each term separately.

The integral of A/x is A*ln|x|, the integral of B/x^2 is -B/x, the integral of C/(x - 1) is C*ln|x - 1|, and the integral of D/(x - 1)^2 is -D/(x - 1).

Finally, combine the integrals of the partial fractions and add the constant of integration, if necessary, to obtain the final result.