Integral from 0 to 3 dxx2x2Solution Use the method of partia

Integral from 0 to 3: dx/x^2-x-2

Use the method of partial fraction decomposition: x^2-x-2=(x-2)(x+1) So 1/(x^2-x-2)= A/(x-2)+B/(x+1), and then use algebra to find A and B. A(x+1)+B(x-2)=1 (A+B)x+(A-2B)=1 Since there are no x\'s on the rhs, A+B=0. And also A-2B=1. A=1+2B so 1+2B+B=0, B=-1/3, A=1/3 Now you can do this indefinite integral easily! (it\'s natural logarithm) BUT, notice that the original argument is undefined at x=2. Because it is a vertical asymptote, it assures you that you will have UNDEFINED answer. The answer is undefined.

$Integral from 0 to 3: dx/x^2-x-2Solution Use the method of partial fraction decomposition: x^2-x-2=(x-2)(x+1) So 1/(x^2-x-2)= A/(x-2)+B/(x+1), and then use alge$

Integral from 0 to 3 dxx2x2Solution Use the method of partia

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