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afcwestwarrior
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Homework Statement
∫(2t)/(t-3)^2
the integral is 2 to 0
ok does it = A/ t-3 + B/(t-3)^2
I'm not sure if you break up (t-3)^2
afcwestwarrior said:Homework Statement
∫(2t)/(t-3)^2
the integral is 2 to 0
ok does it = A/ t-3 + B/(t-3)^2
I'm not sure if you break up (t-3)^2
afcwestwarrior said:So would it be 2t= A(t-3)+ B
=At-3A+B
A(t) - (3A+B)
A=2
3A+B=0
Is this correct.
You have a typo. It's a 3 not a 2.afcwestwarrior said:A=2
and B=6
then would it be 2/ t-2 + 6/ (t-2)^2skipping u substitution part
= 2 ln[t-2] + 6 ln [(t-2)^2
is that correct
afcwestwarrior said:A=3 is what your saying, I'm not following. Where's my mistake
This part is incorrect. What is the anti-derivative of [tex]\frac{1}{(t-3)^2}[/tex]?afcwestwarrior said:oh ok I understand. so would it be 2 Ln [t-3] + 6 ln[(t-3)^2]
There are several techniques of integration including substitution, integration by parts, partial fractions, trigonometric substitution, and completing the square.
To solve a partial fractions problem, you first need to factor the denominator of the rational function into linear or quadratic factors. Then, you set up and solve a system of equations to find the unknown coefficients of each fraction. Finally, you integrate each individual fraction and combine them to get the final answer.
Partial fractions is most commonly used when the denominator of a rational function cannot be factored further, and it contains multiple distinct linear or quadratic factors.
Yes, when writing the partial fractions, the highest degree term should be written first, followed by the next highest degree term, and so on. The constants should be written last.
Yes, partial fractions can be used for improper fractions. In this case, after setting up the system of equations, you may end up with a polynomial of a higher degree on the numerator. This can be solved by performing long division to get a proper fraction, and then using partial fractions to integrate.