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

the book answer was
$$\dfrac{4}{5}\ln{2}+\dfrac{1}{5}\ln {3} =\dfrac{1}{5}\ln{48}$$  Reply With Quote

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 $x=-4$ ...

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

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

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

OK I see

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

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  Reply With Quote

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  Reply With Quote

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281, a=2, b=3, fractions, integral  Posting Permissions

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