281 7.5.9 int with partail fractions

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In summary, the integral is $\displaystyle \int_2^4$ and the partial fractions are $W|A returned these partial fractions but I don't know where the the A=2 B=3 and 5 came from$.
  • #1
karush
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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
$$\dfrac{4}{5}\ln{2}+\dfrac{1}{5}\ln {3}
=\dfrac{1}{5}\ln{48}$$
 
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  • #2
Re: 281 7.5.9 int wiht partail fractions

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

$ \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 $x=-4$ ...

$-4+2 = A(-4-1) + B(-4+4) \implies A = \dfrac{2}{5}$\(\displaystyle \int \dfrac{2}{5(x+4)} + \dfrac{3}{5(x-1)} \, dx = \dfrac{2}{5}\ln(x+4) + \dfrac{3}{5}\ln(x-1) = \dfrac{\ln[(x+4)^2(x-1)^3]}{5}\)$\bigg[ \dfrac{\ln[(x+4)^2(x-1)^3]}{5} \bigg]_2^4 = \dfrac{\ln(64 \cdot 27) - \ln(36)}{5} = \dfrac{\ln(48)}{5}$
 
  • #3
Re: 281 7.5.9 int wiht partail fractions

OK I see

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

:cool:
 
  • #4
Re: 281 7.5.9 int wiht partail fractions

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
 
  • #5
Re: 281 7.5.9 int wiht partail fractions

karush said:
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
 
  • #6
Re: 281 7.5.9 int wiht partail fractions

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

-Dan

Well I miss the pre transistor days...
 

FAQ: 281 7.5.9 int with partail fractions

1. What is a partial fraction?

A partial fraction is a mathematical expression that represents a rational function as a sum of simpler fractions. It is used to simplify complex expressions and make them easier to solve.

2. How do you convert a decimal number to a fraction?

To convert a decimal number to a fraction, you can use the following steps:

  1. Write down the decimal number as a fraction with a denominator of 1 (e.g. 7.5 = 7.5/1).
  2. Multiply both the numerator and denominator by 10 until the decimal point disappears (e.g. 7.5/1 becomes 75/10).
  3. Simplify the fraction if possible (e.g. 75/10 simplifies to 15/2).

3. What is the process for solving an equation with partial fractions?

The process for solving an equation with partial fractions involves the following steps:

  1. Factor the denominator of the rational function into linear factors.
  2. Write the rational function as a sum of partial fractions.
  3. Set up a system of equations by equating the coefficients of the partial fractions to the coefficients of the original rational function.
  4. Solve the system of equations to find the values of the unknown coefficients.
  5. Substitute the values of the coefficients back into the partial fractions and simplify the resulting expression.

4. Can partial fractions be used to solve any type of equation?

No, partial fractions can only be used to solve equations that involve rational functions with distinct linear factors in the denominator. They cannot be used to solve equations with quadratic or higher order factors in the denominator.

5. How can partial fractions be applied in real-world situations?

Partial fractions can be applied in various real-world situations, such as in chemical reactions, electrical circuits, and economics. In chemistry, partial fractions can be used to determine the concentration of a solution. In electrical circuits, they can be used to analyze the behavior of complex networks. In economics, they can be used to model supply and demand curves.

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