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

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Answer 1 · 2018-04-15T05:59:26

$4ln|(x-1)/x|+3/x+C, or, ln(1-1/x)^4+3/x+C$ .

Here is a way to find the Integral $I=int(x+3)/{x^2(x-1)}dx$ ,

without using partial fractions.

$I=int(x+3)/{x^2(x-1)}dx$ ,

$=int{x/{x^2(x-1)}+3/{x^2(x-1)}}dx$ ,

$=int[1/{x(x-1)}+3*(x-(x-1)}/{x^2(x-1)}]dx$ ,

$=int[1/{x(x-1)}+3{x/{x^2(x-1)}-(x-1)/{x^2(x-1)}}]dx$ ,

$=int[1/{x(x-1)}+3/{x(x-1)}-3/x^2]dx$ ,

$=int4/{x(x-1)}dx-3int1/x^2dx$ ,

$=4int{x-(x-1)}/{x(x-1)}dx--3*x^(-2+1)/(-2+1)$ ,

$=4int{1/(x-1)-1/x}dx+3/x$ ,

$=4{ln|(x-1)|-ln|x|}+3/x$ .

$rArr I=4ln|(x-1)/x|+3/x+C, or, ln(1-1/x)^4+3/x+C$ .

Enjoy Maths.!

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