Thread: Integral by partial fractions

I need some help solving this. I know the next step is to multiply out the right side and then collect all like terms but I do not know where to start or how to do this. Supposedly I cannot just pick random values to find A - D.

Originally Posted by gammaman

Supposedly I cannot just pick random values to find A - D.

I'm sorry, but I don't understand what this means...?

Originally Posted by gammaman

To solve for A, B, C, and D, try picking x-values that make your life easier. In this case, an obvious choice would be x = 4, giving you:

16 - 116 + 5 = 0 + B(16 + 3) + 0

-95 = 19(B)

-5 = B

There are no other x-values which will zero out any of the factors, but you can pick some other "nice" values, such as x = 0:

5 = A(-4)(3) - 5(3) + (D)(16)

5 = -12A - 15 + 16D

20 = -12A + 16D

Pick a couple other values, and thus obtain a system of three equations in three unknowns. Solve the system for A, C, and D.

According to my prof, In this example it is not possible to choose values for x and then solve like you just did. According to him we must multiply out the right hand side and combine like terms.

Originally Posted by gammaman

According to my prof, In this example it is not possible to choose values for x and then solve like you just did. According to him we must multiply out the right hand side and combine like terms.

Then do it the way your professor taught you.

By equating coefficients, you'll end up with a system of equations, which you can then solve for the required values.

Originally Posted by gammaman

According to my prof, In this example it is not possible to choose values for x and then solve like you just did. According to him we must multiply out the right hand side and combine like terms.

Like the above poster said then, do it the way your professor wants because he will be the one grading you. However, just so you know, it is perfectly ok to set specific values of x to solve for your other variables. Since you have an equation that is claimed to be true for all x, it certainly holds true for a particular x.