Is my homework solution correct?

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Homework Statement
Typeset problem 6(a): determine whether the integral from 0 to 4 of 1/(x^2 - 3x - 4) dx converges or diverges, and evaluate it if convergent.



3The attempt at a solution

Handwritten solution to an improper integral of 1/(x^2 - 3x - 4) from 0 to 4, partial fractions to A = -1/5 and B = 1/5, concluding the integral diverges to minus infinity.


Is this correct?
 
Do I need to show that I'm approaching 4 from the right?
 
temaire said:
Do I need to show that I'm approaching 4 from the right?

I would say you are approaching 4 from the left. t<4, correct? Why would you want to approach from the right?
 
Dick said:
I would say you are approaching 4 from the left. t<4, correct? Why would you want to approach from the right?

Because if I were approaching ln|x-4| from the right, the graph goes to -∞. How could you approach it from the left?

Or am I supposed to approach the original graph of 1/(x^2 -3x -4) from the left?
 
The problem with your integral is at x=4. To resolve it as an improper integral you want to integrate from x=0 to x=4-epsilon where epsilon>0. That means you are approaching the upper limit from the left. You don't care what the limit is from the right.
 
Dick said:
The problem with your integral is at x=4. To resolve it as an improper integral you want to integrate from x=0 to x=4-epsilon where epsilon>0. That means you are approaching the upper limit from the left. You don't care what the limit is from the right.

So even though the limit is approaching 4 from the left at ln|x-4|, we're infact evaluating the limit as it approaches 4 from the left of 1/(x^2-3x-4)?
 
temaire said:
So even though the limit is approaching 4 from the left at ln|x-4|, we're infact evaluating the limit as it approaches 4 from the left of 1/(x^2-3x-4)?

Now you are just confusing me. You are approaching x=4 from the left. Period. Approaching ln|x-4| from the left gives you the behavior of the integral of 1/(x^2-3x-4) on the interval [0,4].
 
This is what I was confused about.

Graph of y = ln|x-4| on a grid with axes, showing a curve that drops steeply toward a vertical asymptote at x = 4, with a red arrow pointing down along that asymptote.


In the graph above, you can approach 4 from the right but not from the left.
 
It's an absolute value, temaire. |x-4|. Doesn't that mean anything to you? :)
 
Oh, so this is the graph. (x is from 3 to 5)

Graph of a curve dipping to a sharp cusp at y = -4 near x = 4, with the curve rising on both sides and crossing the x-axis near x = 3 and x = 5; axis ticks labeled 3.5, 4, 4.5, 5 on x and -1 to -4 on y.


I understand now.