How do you integrate $\int \frac{x - 5}{{(x - 2)}^{2}}$ $int ( x-5)/(x-2)^2$ using partial fractions?

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How do you integrate $\int \frac{x - 5}{{(x - 2)}^{2}}$ $int ( x-5)/(x-2)^2$ using partial fractions?

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Answer 1 · 2016-11-02T20:09:52

The answer is $= ln (x - 2) + \frac{3}{x - 1} + C$ $=ln(x-2)+3/(x-1)+C$

Explanation:

Let's start with the decomposition of the partial fractions
$= \frac{A + B (x - 2)}{{(x - 2)}^{2}}$ $=(A+B(x-2))/(x-2)^2$
$:.$ $x-5=A+B(x-2)$
Coefficients of x $B=1$
and $-5=A-2B$ $=>$ $A=-3$
$(x-5)/(x-2)^2=-3/(x-2)^2+1/(x-2)$
So $int((x-5)dx)/(x-2)^2=intdx/(x-2)-int(3dx)/(x-2)^2$
$=ln(x-2)+3/(x-1)+C$

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How do you integrate $\int \frac{x - 5}{{(x - 2)}^{2}}$ $int ( x-5)/(x-2)^2$ using partial fractions?

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How do you integrate $\int \frac{x - 5}{{(x - 2)}^{2}}$ $int ( x-5)/(x-2)^2$ using partial fractions?