How do you find int(x-1) / ( 2x^2 - 3x -3) dx∫x−12x2−3x−3dx using partial fractions?

1 Answer
ali ergin
Jun 3, 2016

int (x-1)/(2x^2-3x-3) d x=1/4 l n|2x^2-3x-3|-1/4 1/sqrt33 l n|(4x-3-sqrt33)/(4x-3+sqrt 33)|+C∫x−12x2−3x−3dx=14ln∣∣2x2−3x−3∣∣−141√33ln∣∣∣4x−3−√334x−3+√33∣∣∣+C

Explanation:

int (x-1)/(2x^2-3x-3) d x=?∫x−12x2−3x−3dx=?

2x^2-3x-3=u" ; "d u=(4x-3)*d x 2x2−3x−3=u ; du=(4x−3)⋅dx

=1/4int (4(x-1))/(2x^2-3x-3) d x =14∫4(x−1)2x2−3x−3dx

=1/4[ int (4x-3)/(2x^2-3x-3) d x-1/(2x^2-3x-3) d x]=14[∫4x−32x2−3x−3dx−12x2−3x−3dx]

=1/4[int (d u)/u-(d x)/(2x^2-3x-3)]=14[∫duu−dx2x2−3x−3]

Delta=sqrt(b^2-4*a*c)

Delta=sqrt((-3)^2+4*2*3)=sqrt(33)

=1/4 l n|u|-1/4int (d x)/(2x^2-3x-3)

=1/4 l n| 2x^2-3x-3|-1/4int (d x)/(2x^2-3x-3)

=1/4 l n|2x^2-3x-3|-1/4 1/Delta l n|(2ax+b-Delta)/(2ax+b+Delta)|+C

2ax=2*2*x=4x" ;"b=-3" ; "Delta =sqrt33

=1/4 l n|2x^2-3x-3|-1/4 1/sqrt33 l n|(4x-3-sqrt33)/(4x-3+sqrt 33)|+C

int (x-1)/(2x^2-3x-3) d x=1/4 l n|2x^2-3x-3|-1/4 1/sqrt33 l n|(4x-3-sqrt33)/(4x-3+sqrt 33)|+C

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