Thread: 281 7.5.9 int wiht partail fractions

1. Evaluate the integral.
$$\displaystyle \int_2^4 \dfrac{x+2}{x^2+3x-4}\, dx$$
W|A returned these partial fractions but I don't know where the the A=2 B=3 and 5 came from
$$\dfrac{x+2}{(x+4)(x-1)} =\dfrac{2}{5(x+4)}+\dfrac{3}{5(x-1)}$$

$$\dfrac{4}{5}\ln{2}+\dfrac{1}{5}\ln {3} =\dfrac{1}{5}\ln{48}$$

2.

3. $\dfrac{x+2}{x^2+3x-4} = \dfrac{x+2}{(x+4)(x-1)} = \dfrac{A}{x+4} + \dfrac{B}{x-1}$

$\dfrac{x+2}{(x+4)(x-1)} = \dfrac{A(x-1)}{(x+4)(x-1)} + \dfrac{B(x+4)}{(x+4)(x-1)}$

common denominator for all terms $\implies$ numerators form the equation ...

$x+2 = A(x-1) + B(x+4)$

two methods ...

(1) matching coefficients

$x+2 = (A+B)x + (4B-A) \implies A+B = 1 \text{ and } 4B-A = 2$

solving the system by elimination yields $5B = 3 \implies B = \dfrac{3}{5} \text{ and } A = \dfrac{2}{5}$

$\dfrac{x+2}{(x+4)(x-1)} = \dfrac{2}{5(x+4)} + \dfrac{3}{5(x-1)}$

(2) Heaviside method ... pick strategic values for $x$ to eliminate $A$ or $B$

$x+2 = A(x-1) + B(x+4)$

let $x=1$ ...

$1+2 = A(1-1) + B(1+4) \implies B = \dfrac{3}{5}$

let $x=-4$ ...

$-4+2 = A(-4-1) + B(-4+4) \implies A = \dfrac{2}{5}$

$\displaystyle \int \dfrac{2}{5(x+4)} + \dfrac{3}{5(x-1)} \, dx = \dfrac{2}{5}\ln(x+4) + \dfrac{3}{5}\ln(x-1) = \dfrac{\ln[(x+4)^2(x-1)^3]}{5}$

$\bigg[ \dfrac{\ln[(x+4)^2(x-1)^3]}{5} \bigg]_2^4 = \dfrac{\ln(64 \cdot 27) - \ln(36)}{5} = \dfrac{\ln(48)}{5}$

OK I see

you let on of the variables be multiplied by zero to isolate the the other one

wow with 377 views that was a great help

they did one in class but I got lost on getting A and B

really appreciate the steps

if this were yahoo it would be torture to read without the latex

Mahalo

6. Originally Posted by karush
wow with 377 views that was a great help

they did one in class but I got lost on getting A and B

really appreciate the steps

if this were yahoo it would be torture to read without the latex

Mahalo
Yeah but I still sometimes miss those old days where DOS was high tech and Windows hadn't been invented yet.

-Dan