How do you find $int (3x^2 - 10) / (x^2-4x+4) dx$ using partial fractions?

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Answer 1 · 2016-02-27T17:17:04

$3x+12lnabs(x-2)-2/(x-2)+C$

First, since the degrees of the numerator and denominator are equal, use polynomial long division to rewrite the expression:

$(3x^2-10)/(x^2-4x+4)=3+(12x-22)/(x^2-4x+4)$

Now, perform partial fraction decomposition on $(12x-22)/(x^2-4x+4)$ , recognizing that $x^2-4x+4=(x-2)^2$ .

$(12x-22)/((x-2)^2)=A/(x-2)+B/(x-2)^2$

Note that since the term is squared, it will be repeated.

Multiply both sides by $(x-2)^2$ to see that

$12x-22=A(x-2)+B$

When we set $x=2$ , we see that

$12(2)-22=A(0)+B$

$2=B$

Arbitrarily, set $x=3$ to solve for $A$ , recalling that $B=2$ :

$12(3)-22=A(1)+2$

$A=12$

Thus,

$(3x^2-10)/(x^2-4x+4)=3+12/(x-2)+2/(x-2)^2$

Now, we can integrate more simply:

$int3+12/(x-2)+2/(x-2)^2dx$

$=3x+12lnabs(x-2)-2/(x-2)+C$

How do you find int (3x^2 - 10) / (x^2-4x+4) dxint (3x^2 - 10) / (x^2-4x+4) dx using partial fractions?