How do you integrate int (4x-2) /( 3(x-1)^2)∫4x−23(x−1)2 using partial fractions?

1 Answer
Shwetank Mauria
May 12, 2016

int(4x-2)/(3(x-1)^2)dx=4/3ln(x-1)-4/(3(x-1))+c∫4x−23(x−1)2dx=43ln(x−1)−43(x−1)+c

Explanation:

Let us first find partial fractions of (4x-2)/(3(x-1)^2)4x−23(x−1)2 and for this let

(4x-2)/((x-1)^2)hArrA/(x-1)+B/(x-1)^24x−2(x−1)2⇔Ax−1+B(x−1)2 or

(4x-2)/((x-1)^2)hArr(A(x-1)+B)/((x-1)^2)=(Ax+B-A)/((x-1)^2)4x−2(x−1)2⇔A(x−1)+B(x−1)2=Ax+B−A(x−1)2

Hence A=4A=4 and B-A=-2B−A=−2 i.e. B=4-2=2B=4−2=2

Hence int(4x-2)/(3(x-1)^2)dx=1/3int[2/(x-1)+2/(x-1)^2]dx∫4x−23(x−1)2dx=13∫[2x−1+2(x−1)2]dx

= 2/3int2/(x-1)dx+2/3int2/(x-1)^2dx+k23∫2x−1dx+23∫2(x−1)2dx+k

= 4/3ln(x-1)-4/(3(x-1))+c43ln(x−1)−43(x−1)+c

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