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

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The discussion revolves around integrating the function (x + 4)/(x^2 + 5x - 6) using partial fractions. The correct factorization of the denominator is identified as (x - 1)(x + 6), leading to the setup of A and B for the partial fractions. The values of A and B are calculated, but there are errors in the initial solutions presented. The final integration results in a logarithmic expression that can be simplified using logarithmic properties to match the provided solution. The importance of correctly applying these properties and the constant of integration is emphasized.
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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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beaf123 said:

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 \frac{x+4}{x^2+5x-6}

How did you get A+B=1 ?
 
Okey, thanks.

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

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

Ax + Bx = xA+B =1
 
beaf123 said:
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
 
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
 
beaf123 said:
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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