What is $int (x^3-2x^2+6x-3 ) / (-x^2+ 9 x +2 )$ ?

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Answer 1 · 2016-03-13T03:30:40

$int(x^3-2x^2+6x-3)/(-x^2+9x+2)dx=$
$-x^2/2-7x-71/2*ln(x^2-9x-2)$
$-(661sqrt(89))/178*ln((2x-9-sqrt89)/(2x-9+sqrt89))+C$

From the given integrand, divide the numerator by the denominator. So that
$(x^3−2x^2+6x−3)/(−x^2+9x+2)=-x-7-(71x+11)/(x^2-9x-2)$

We have to simplify the third term, so that

$(71x+11)/(x^2-9x-2)=((71/2)(2x-9+9+(2*11)/71))/(x^2-9x-2)$

that is to force into the numerator a term $(2x-9)$ which is the derivative of $(x^2-9x-2)$ , enabling us to use $int((du)/u)=ln u+C$ .

We can also use $int (du)/(u^2-a^2)=1/(2a)*ln((u-a)/(u+a))+C$ afterwards

the continuation is

$(71x+11)/(x^2-9x-2)=((71/2)(2x-9+661/71))/(x^2-9x-2)$

$(71x+11)/(x^2-9x-2)=((71/2)(2x-9)+(71/2)(661/71))/(x^2-9x-2)$

Now, we are ready to integrate

$int(x^3−2x^2+6x−3)/(−x^2+9x+2)dx=int(-x-7-(71x+11)/(x^2-9x-2))dx$

$int(x^3−2x^2+6x−3)/(−x^2+9x+2)dx=$

$int(-x-7-((71/2)(2x-9)+(71/2)(661/71))/(x^2-9x-2))dx$

$int(-x-7-((71/2)(2x-9))/(x^2-9x-2)-((71/2)(661/71))/(x^2-9x-2))dx$

also by "completing the square"

$int(-x-7-((71/2)(2x-9))/(x^2-9x-2)-((661/2))/((x-9/2)^2-(sqrt89/2)^2))dx$

and

$-x^2/2-7x$
$-71/2*ln(x^2-9x-2)-(661sqrt(89))/178*ln((2x-9-sqrt89)/(2x-9+sqrt89))+C$

God bless....I hope the explanation is useful.

What is int (x^3-2x^2+6x-3 ) / (-x^2+ 9 x +2 )int (x^3-2x^2+6x-3 ) / (-x^2+ 9 x +2 )?