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

Harlen Pritchard 2020-10-25 Answered
Evaluate the integral. 01(x4)(x25x+6dx)
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Expert Answer

Mayme
Answered 2020-10-26 Author has 103 answers
01x4x25x+6dx
01x4x23x2x+6dx
01x4[x23x]+[2x+6]dx
01x4x[x3x]2[x3]dx
01x4(x2)(x3)dx We have,
x4(x2)(x3)=Ax2+Bx3
x4(x2)(x3)=A(x3)(x2)(x3)+B(x2)(x2)(x3)
x4(x2)(x3)=A(x3)+B(x2)(x2)(x3)
On comparing the numerators, we get
x4=A(x3)+B(x2) (1)
Substitute x=2 in Equ(1), To get
24=A(23)+B(22)
2=AA=2
x4=A(x3)+B(x2) (1)
Substitute x=3 in Eqn(1), To get
34=A(33)+B(32)
1=B
01x4(x2)(x3)dx=012x21x3dx
=[2ln|x2|ln|x3|]01
=[2ln|12|ln|13|][2ln|02|ln|03|]
=[0ln2][2ln2ln3]=ln33ln2
=ln3ln23ln8=ln(38)
Remember that: lnxlny=ln(xy)
Also Remember that:alnx=lnxa
Result
01x4x25x+6dx=ln33ln2=ln(38)
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