How do you integrate $int (5x-1)/(x^2-x-2)$ using partial fractions?

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Answer 1 · 2016-05-13T04:25:18

$2 ln (x+1) + 3 ln (x-2) + c$ , with $x > 2$

Let $f(x)=(5x-1)/(x^2-x-2)$

$=(5x-1)/((x-2)(x+1))$

$=(A(x-2)+B(x+1))/((x-2)(x+1))$

$=A/(x+1)+B/(x-2)$

Determine matching A and B from $A(x-2)+B(x+1)=5x-1$

Comparing coefficients of x, A+B=5. Comparing constants, $-2A+B=-1$ .

Solving, A = 2 and B = 3.

Now, $intf(x) dx$

$= 2 int1/(x+1) dx + 3 int 1/(x-2) dx$

$=2 ln (x+1)+ 3 ln (x-2) + C$

For both the logarithms to be real, $x > 2$

How do you integrate int (5x-1)/(x^2-x-2)int (5x-1)/(x^2-x-2) using partial fractions?