Conceptual Integration by Parts problem
- Thread starter Jaskaran
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Jaskaran said:
Can be broken as
How do I solve for A and B ?!! :?
x[sup:239zvrgy]2[/sup:239zvrgy] + 3x - 40 ? factorize it then use "partial fraction" techniques or technology (as using a CAS calculator)
Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.
If you factor the denominator, you get
\(\displaystyle (x-5)(x+8)\)
This can then be expanded into Partial fractions:
\(\displaystyle \frac{A}{x-5}+\frac{B}{x+8}=10x-102\)
\(\displaystyle A(x+8)+B(x-5)=10x-102\)
Distribute to eliminate parentheses on the left, equate coefficients, and solve for A and B.
You can also do partial fractions by subbing in the value of x that makes x+8=0
Then, sub in the value that makes x-5=0.
Some prefer this method over the other.
For instance, if x=-8:
\(\displaystyle A(-8+8)+B(-8-5)=10(-8)-102\)
There. You have B.
Do the same for A and you have it.
\(\displaystyle (x-5)(x+8)\)
This can then be expanded into Partial fractions:
\(\displaystyle \frac{A}{x-5}+\frac{B}{x+8}=10x-102\)
\(\displaystyle A(x+8)+B(x-5)=10x-102\)
Distribute to eliminate parentheses on the left, equate coefficients, and solve for A and B.
You can also do partial fractions by subbing in the value of x that makes x+8=0
Then, sub in the value that makes x-5=0.
Some prefer this method over the other.
For instance, if x=-8:
\(\displaystyle A(-8+8)+B(-8-5)=10(-8)-102\)
There. You have B.
Do the same for A and you have it.
Looks like that this problem is more related to integration of rational functions, not integration by parts. See some other typical examples at http://www.math24.net/integration-of-rational-functions.html.