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

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Answer 1 · 2016-05-13T04:25:18

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

Let $f (x) = \frac{5 x - 1}{x^{2} - x - 2}$ $f(x)=(5x-1)/(x^2-x-2)$

$= \frac{5 x - 1}{(x - 2) (x + 1)}$ $=(5x-1)/((x-2)(x+1))$

$= \frac{A (x - 2) + B (x + 1)}{(x - 2) (x + 1)}$ $=(A(x-2)+B(x+1))/((x-2)(x+1))$

$= \frac{A}{x + 1} + \frac{B}{x - 2}$ $=A/(x+1)+B/(x-2)$

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

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

Solving, A = 2 and B = 3.

Now, $\int f (x) d x$ $intf(x) dx$

$= 2 \int \frac{1}{x + 1} d x + 3 \int \frac{1}{x - 2} d x$ $= 2 int1/(x+1) dx + 3 int 1/(x-2) dx$

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

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

How do you integrate ∫5x−1x2−x−2int (5x-1)/(x^2-x-2) using partial fractions?