How do you integrate x^2/((x-3)^2(x+4))x2(x3)2(x+4) using partial fractions?

2 Answers
Jun 21, 2018

intx^2/((x-3)^2(x+4))dx=16/49ln|x+4|+33/49ln|x-3|-9/(7(x-3))+Cx2(x3)2(x+4)dx=1649ln|x+4|+3349ln|x3|97(x3)+C

Explanation:

x^2/((x-3)^2(x+4))=A/(x+4)+B/(x-3)+C/(x-3)^2x2(x3)2(x+4)=Ax+4+Bx3+C(x3)2
x^2=A(x-3)^2+B(x-3)(x+4)+C(x+4)x2=A(x3)2+B(x3)(x+4)+C(x+4)

Substituting x=3x=3:
9=7C9=7C so C=9/7C=97

Substituting x=-4x=4:
16=49A16=49A so A=16/49A=1649

Looking at the coefficients of x^2x2 on both sides, we see that x^2=Ax^2+Bx^2x2=Ax2+Bx2 or that A+B=1A+B=1.
B=1-16/49=33/49B=11649=3349

Therefore, the partial fraction decomposition of x^2/((x-3)^2(x+4))x2(x3)2(x+4) is (16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^21649x+4+3349x3+97(x3)2.

intx^2/((x-3)^2(x+4))dx=int((16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^2)dxx2(x3)2(x+4)dx=(1649x+4+3349x3+97(x3)2)dx

int((16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^2)dx=16/49ln|x+4|+33/49ln|x-3|-9/(7(x-3))+C(1649x+4+3349x3+97(x3)2)dx=1649ln|x+4|+3349ln|x3|97(x3)+C

Jun 21, 2018

int (x^2dx)/((x-3)^2(x+4)) =33/49ln abs(x-3)-9/(7(x-3)) +16/49ln abs(x+4)+Cx2dx(x3)2(x+4)=3349ln|x3|97(x3)+1649ln|x+4|+C

Explanation:

Apply partial fraction decomposition:

x^2/((x-3)^2(x+4)) = A/(x-3)+B/(x-3)^2 +C/(x+4)x2(x3)2(x+4)=Ax3+B(x3)2+Cx+4

x^2/((x-3)^2(x+4)) = (A(x-3)(x+4)+B(x+4)+C(x-3)^2)/((x-3)^2(x+4))x2(x3)2(x+4)=A(x3)(x+4)+B(x+4)+C(x3)2(x3)2(x+4)

x^2 = A(x^2+x-12)+Bx+4B+C(x^2-6x+9))x2=A(x2+x12)+Bx+4B+C(x26x+9))

x^2 = Ax^2+ Ax-12A +Bx+4B +Cx^2 -6Cx+9Cx2=Ax2+Ax12A+Bx+4B+Cx26Cx+9C

x^2 = (A+C)x^2 + (A+B-6C)x - (12A-4B-9C)x2=(A+C)x2+(A+B6C)x(12A4B9C)

{(A+C=1),(A+B-6C=0),(12A-4B-9C=0):}

{(A=1-C),(B-7C=-1),(4B+21C=12):}

{(A=1-C),(-4B+28C=4),(4B+21C=12):}

{(A=33/49),(B=9/7),(C=16/49):}

Then:

int (x^2dx)/((x-3)^2(x+4)) =33/49int dx /(x-3)+9/7int dx/(x-3)^2 +16/49int dx/(x+4)

int (x^2dx)/((x-3)^2(x+4)) =33/49ln abs(x-3)-9/(7(x-3)) +16/49ln abs(x+4)+C