
MHB Master
#1
October 8th, 2019,
15:29
Evaluate the integral.
$$\displaystyle \int_2^4
\dfrac{x+2}{x^2+3x4}\, dx$$
WA returned these partial fractions but I don't know where the the A=2 B=3 and 5 came from
$$
\dfrac{x+2}{(x+4)(x1)}
=\dfrac{2}{5(x+4)}+\dfrac{3}{5(x1)}$$
the book answer was
$$\dfrac{4}{5}\ln{2}+\dfrac{1}{5}\ln {3}
=\dfrac{1}{5}\ln{48}$$
Last edited by karush; October 8th, 2019 at 15:50.

October 8th, 2019 15:29
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MHB Craftsman
#2
October 8th, 2019,
16:34
$\dfrac{x+2}{x^2+3x4} = \dfrac{x+2}{(x+4)(x1)} = \dfrac{A}{x+4} + \dfrac{B}{x1}$
$ \dfrac{x+2}{(x+4)(x1)} = \dfrac{A(x1)}{(x+4)(x1)} + \dfrac{B(x+4)}{(x+4)(x1)}$
common denominator for all terms $\implies$ numerators form the equation ...
$x+2 = A(x1) + B(x+4)$
two methods ...
(1) matching coefficients
$x+2 = (A+B)x + (4BA) \implies A+B = 1 \text{ and } 4BA = 2$
solving the system by elimination yields $5B = 3 \implies B = \dfrac{3}{5} \text{ and } A = \dfrac{2}{5}$
$ \dfrac{x+2}{(x+4)(x1)} = \dfrac{2}{5(x+4)} + \dfrac{3}{5(x1)}$
(2) Heaviside method ... pick strategic values for $x$ to eliminate $A$ or $B$
$x+2 = A(x1) + B(x+4)$
let $x=1$ ...
$1+2 = A(11) + B(1+4) \implies B = \dfrac{3}{5}$
let $x=4$ ...
$4+2 = A(41) + B(4+4) \implies A = \dfrac{2}{5}$
$ \displaystyle \int \dfrac{2}{5(x+4)} + \dfrac{3}{5(x1)} \, dx = \dfrac{2}{5}\ln(x+4) + \dfrac{3}{5}\ln(x1) = \dfrac{\ln[(x+4)^2(x1)^3]}{5}$
$\bigg[ \dfrac{\ln[(x+4)^2(x1)^3]}{5} \bigg]_2^4 = \dfrac{\ln(64 \cdot 27)  \ln(36)}{5} = \dfrac{\ln(48)}{5}$

MHB Master
#3
October 12th, 2019,
15:24
Thread Author
OK I see
you let on of the variables be multiplied by zero to isolate the the other one

MHB Master
#4
Today,
17:16
Thread Author
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

MHB Master
#5
Today,
21:16
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