How do you integrate $int 18/ ( x^2 (x+3) ) dx$ using partial fractions?

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Answer 1 · 2016-02-09T02:26:41

$int18/(x^2(x+3))dx=-2ln|x| -6/x + 2ln|x+3| + C$

$18/(x^2(x+3)) = A/x + B/x^2 + C/(x+3)$

$=> 18 = Ax(x+3) + B(x+3) + Cx^2$

$= (A+C)x^2 + (3A+B)x + 3B$

$=>{(A+C=0),(3A+B=0),(3B=18):}$

$=>{(A=-2),(B=6),(C=2):}$

$=>18/(x^2(x+3)) = -2/x+6/x^2+2/(x+3)$

$=>int18/(x^2(x+3))dx = int-2/x+6/x^2+2/(x+3)dx$

$=-2int1/xdx + 6int1/x^2dx + 2int1/(x+3)dx$

$=-2ln|x| -6/x + 2ln|x+3| + C$

How do you integrate int 18/ ( x^2 (x+3) ) dxint 18/ ( x^2 (x+3) ) dx using partial fractions?