Evaluate the integral. ∫01(x−4)(x2−5x+6dx)

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∫01x−4x2−5x+6dx  ∫01x−4x2−3x−2x+6dx  ∫01x−4[x2−3x]+[−2x+6]dx  ∫01x−4x[x−3x]−2[x−3]dx  ∫01x−4(x−2)(x−3)dx We have,  x−4(x−2)(x−3)=Ax−2+Bx−3  x−4(x−2)(x−3)=A(x−3)(x−2)(x−3)+B(x−2)(x−2)(x−3)  x−4(x−2)(x−3)=A(x−3)+B(x−2)(x−2)(x−3)  On comparing the numerators, we get  x−4=A(x−3)+B(x−2)⇒ (1)  Substitute x=2 in Equ(1), To get  2−4=A(2−3)+B(2−2)  −2=−A⇒A=2  x−4=A(x−3)+B(x−2)⇒ (1)  Substitute x=3 in Eqn(1), To get  3−4=A(3−3)+B(3−2)  −1=B  ∫01x−4(x−2)(x−3)dx=∫012x−2−1x−3dx  =[2ln⁡|x−2|−ln⁡|x−3|]01  =[2ln⁡|1−2|−ln⁡|1−3|]−[2ln⁡|0−2|−ln⁡|0−3|]  =[0−ln⁡2]−[2ln⁡2−ln⁡3]=ln⁡3−3ln⁡2  =ln⁡3−ln⁡23−ln⁡8=ln⁡(38)  Remember that: ln⁡x−ln⁡y=ln⁡(xy)  Also Remember that:a⋅ln⁡x=ln⁡xa  Result  ∫01x−4x2−5x+6dx=ln⁡3−3ln⁡2=ln⁡(38)

Evaluate the integral. int_0^1frac{(x-4)}{(x^2-5x+6dx)}

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