How do you integrate $int (x-3x^2)/((x+3)(x-1)(x-7))$ using partial fractions?

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Answer 1 · 2016-09-27T07:43:24

$=-3/4In(x+3)+1/12In(x-1)-7/3In(x-7)$

first ,ignore the integrate sign and do partial fraction of the function

$(x-3x^2)/((x+3)(x-1)(x-7))=A/(x+3)+B/(x-1)+C/(x-7)$

$x-3x^2=A(x-1)(x-7))+B(x+3)(x-7))+C(x+3)(x-1)$

subtitute $x=-3$

$(-3)-3(-3)^2=A(-3-1)(-3-7)$

$-3-27=A(-4)(-10)$

$-30=40A$

$A=-3/4$

subtitute $x=1$

$(1)-3(1)^2=B(1+3)(1-7)$

$-2=B(4)(-6)$

$-2=-24B$

$B=1/12$

subtitute $x=7$

$(7)-3(7)^2=C(7+3)(7-1)$

$7-147=C(10)(6)$

$-140=60C$

$C=-7/3$

$int(x-3x^2)/((x+3)(x-1)(x-7))$

$=int-3/(4(x+3))+1/(12(x-1))-7/(3(x-7))$

$=-3/4int1/(x+3)+1/12int1/(x-1)-7/3int1/(x-7)$

$=-3/4In(x+3)+1/12In(x-1)-7/3In(x-7)$

How do you integrate int (x-3x^2)/((x+3)(x-1)(x-7)) int (x-3x^2)/((x+3)(x-1)(x-7)) using partial fractions?