Integrate: (x^2 + 4)/ (x^2+5x-6) dx

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  • #1
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Homework Statement



"Express the integrand (what does "integrand" mean?) as a sum of partial fractions and evaluate the integrals.

∫(x + 4)/ (x^2+5x-6) dx

Homework Equations





The Attempt at a Solution



x^2+5x-6 = (x-1)(x+6)

Gives:

∫ A/(x-1) + B/(x+6) dx

Findig A and B:

A(x+6) + B(x-1)

A+B=1

6A-B =4

A= 5/7
B= - (2/7)


Then:

∫ (5/7) (1/x-1) dx +∫ - (3/7)(1/x+6) dx

Gives I think:

5/7ln(x-1) -2/7ln(x+6)


But its wrong accordng to the solution. I can post the solution here if you want me to:
 
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Answers and Replies

  • #2
Mentallic
Homework Helper
3,798
94

Homework Statement



"Express the integrand (what does "integrand" mean?) as a sum of partial fractions and evaluate the integrals.

∫(x + 4)/ (x^2+5x-6) dx

Homework Equations





The Attempt at a Solution



x^2+5x-6 = (x-1)(x+6)

Gives:

∫ A/(x-1) + B/(x+6) dx

Findig A and B:

A(x+6) + B(x-1)

A+B=1

6A-B =4

A= 5/7
B= - (3/7)


Then:

∫ (5/7) (1/x-1) dx +∫ - (3/7)(1/x+6) dx

Gives I think:

5/7ln(x-1) -3/7ln(x+6)


But its wrong accordng to the solution. I can post the solution here if you want me to:
The integrand is the expression being integrated, in this case, the rational function [tex]\frac{x+4}{x^2+5x-6}[/tex]

How did you get A+B=1 ?
 
  • #3
41
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Okey, thanks.

A(x+6) + B(x-1) = x+4

Ax +Bx +6A -B = x+4

Ax + Bx = x


A+B =1
 
  • #4
Mentallic
Homework Helper
3,798
94
Okey, thanks.

A(x+6) + B(x-1) = x+4

Ax +Bx +6A -B = x+4

Ax + Bx = x


A+B =1
Oh I see, you were comparing coefficients :smile:

Sorry I didn't spot your error before, but you incorrectly solved the simultaneous equations in A and B. 1 - 5/7 = 2/7
 
  • #5
41
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Oh, I changed it now. It is correct so far, but I dont understand how they get


5/7ln(x-1) -2/7ln(x+6)


to become ( from solution):


(1/7)ln[(x+6)^2(x-1)^5] + C
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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Oh, I changed it now. It is correct so far, but I don't understand how they get

5/7ln(x-1) -2/7ln(x+6)

to become ( from solution):

(1/7)ln[(x+6)^2(x-1)^5] + C
That should be
5/7ln(x-1) + 2/7ln(x+6)​
Then use properties of logarithms to get the desired result.

Of course the given answer includes the constant of integration.
 

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