partial fractions
- Thread starter refid
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I'm going to guess that you mean the following:
. . . . .Integrate: \(\displaystyle \large{\int{\,\frac{1}{(x\,-\,3)(x\,+\,2)}\,}dx}\)
From your subject line, I will guess that your "solution" (what you say is what you "did" to "solve"), namely, "Ax + Bx - 3B + 2B =", is one step in the process of integration, namely, splitting up the given integrand into two fractions using decomposition by partial fractions.
If so, then you need to recall the reason for setting up the fractions: You are trying to find the fraction sum that equals the given fraction. So set the two equal:
. . . . .\(\displaystyle \large{\frac{1}{(x\,-\,3)(x\,+\,2)}\,=\,\frac{A}{x\,-\,3}\,+\,\frac{B}{x\,+\,2}}\)
There are various methods for solving this equation for the values of A and B. One might be to multiply through by the common denominator to get:
. . . . .1 = A(x + 2) + B(x - 3)
. . . . .(0)x + (1)1 = (A + B)x + (2A - 3B)1
Then equate coefficients. Solve the resulting system for A and B. Then do the integration.
If I have misunderstood your meaning, please reply with corrections. Thank you.
Eliz.
. . . . .Integrate: \(\displaystyle \large{\int{\,\frac{1}{(x\,-\,3)(x\,+\,2)}\,}dx}\)
From your subject line, I will guess that your "solution" (what you say is what you "did" to "solve"), namely, "Ax + Bx - 3B + 2B =", is one step in the process of integration, namely, splitting up the given integrand into two fractions using decomposition by partial fractions.
If so, then you need to recall the reason for setting up the fractions: You are trying to find the fraction sum that equals the given fraction. So set the two equal:
. . . . .\(\displaystyle \large{\frac{1}{(x\,-\,3)(x\,+\,2)}\,=\,\frac{A}{x\,-\,3}\,+\,\frac{B}{x\,+\,2}}\)
There are various methods for solving this equation for the values of A and B. One might be to multiply through by the common denominator to get:
. . . . .1 = A(x + 2) + B(x - 3)
. . . . .(0)x + (1)1 = (A + B)x + (2A - 3B)1
Then equate coefficients. Solve the resulting system for A and B. Then do the integration.
If I have misunderstood your meaning, please reply with corrections. Thank you.
Eliz.
umm... i got stumped again .....
. . . . .Integrate: \(\displaystyle \large{\int{\,\frac{x^2+1}{(x\,+\,1)(x\,+\,2)}\,}dx}\)
I tried using the above method of assigning A and B it didnt come out to the correct answers, how would I approach this question?
. . . . .Integrate: \(\displaystyle \large{\int{\,\frac{x^2+1}{(x\,+\,1)(x\,+\,2)}\,}dx}\)
I tried using the above method of assigning A and B it didnt come out to the correct answers, how would I approach this question?
Hello, refid!
They snuck another one by you . . .
\(\displaystyle \;\;\;\)(But I'm not laughing . . . I've been thre.)
\(\displaystyle \;\;\;\)It's an "improper fraction" ... Some long division is required . . .
\(\displaystyle \L\int\frac{x^2\,+\,1}{x^2\,+\,3x\,+\,2}\,dx \;= \;\int\left[1\,-\,\frac{3x\,+\,1}{(x\,+\,1)(x\,+\,2)}\right]\,dx\)
Got it?
They snuck another one by you . . .
\(\displaystyle \;\;\;\)(But I'm not laughing . . . I've been thre.)
Note that the numerator and denominator are both quadratics.
\(\displaystyle \;\;\;\)It's an "improper fraction" ... Some long division is required . . .
\(\displaystyle \L\int\frac{x^2\,+\,1}{x^2\,+\,3x\,+\,2}\,dx \;= \;\int\left[1\,-\,\frac{3x\,+\,1}{(x\,+\,1)(x\,+\,2)}\right]\,dx\)
Got it?
Sometimes long division isn't necessary, if you can make some good guesses. For example, one way to reduce the integrand (without long division) would be:
\(\displaystyle \L\qquad\begin{eqnarray*}
\frac{x^2+1}{(x+1)(x+2)}
&=&\frac{x^2+3x+2-(3x+1)}{(x+1)(x+2)}\\
&=&\frac{(x+1)(x+2)-(3x+1)}{(x+1)(x+2)}\\
&=&\frac{(x+1)(x+2)}{(x+1)(x+2)}-\frac{3x+1}{(x+1)(x+2)}\\
&=&1-\frac{3x+1}{(x+1)(x+2)}
\end{eqnarray*}\)
Of course, it wasn't really a guess we made in the first line. The motivation is to make \(\displaystyle (x+1)(x+2)=x^2+3x+2\) appear somewhere the numerator, which would enable us to cancel it out later (and that happens between the third and fourth lines).
\(\displaystyle \L\qquad\begin{eqnarray*}
\frac{x^2+1}{(x+1)(x+2)}
&=&\frac{x^2+3x+2-(3x+1)}{(x+1)(x+2)}\\
&=&\frac{(x+1)(x+2)-(3x+1)}{(x+1)(x+2)}\\
&=&\frac{(x+1)(x+2)}{(x+1)(x+2)}-\frac{3x+1}{(x+1)(x+2)}\\
&=&1-\frac{3x+1}{(x+1)(x+2)}
\end{eqnarray*}\)
Of course, it wasn't really a guess we made in the first line. The motivation is to make \(\displaystyle (x+1)(x+2)=x^2+3x+2\) appear somewhere the numerator, which would enable us to cancel it out later (and that happens between the third and fourth lines).
- Joined
- Feb 4, 2004
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To check an integration, differentiate and see if you get what you started with.refid said:
. . . . .f(x) = x - 5 ln(x + 2) - 2 ln(x + 1) + C
. . . . .f'(x) = 1 - 5/(x + 2) - 2/(x + 1)
. . . . .. . . .= [(x + 2)(x + 1) - 5(x + 1) - 2(x + 2)] / [(x + 2)(x + 1)]
. . . . .. . . .= [x<sup>2</sup> + 2x + 3x + 2 - 5x - 5 - 2x - 4] / [(x + 2)(x + 1)]
. . . . .. . . .= [x<sup>2</sup> - 3x - 7] / [(x + 2)(x + 1)]
Is that what you started with?
Eliz.
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1) Practice.refid said:
2) Don't be afraid to try more than one thing. Just because the first technique didn't work, doesn't mean the next one won't.
3) Practice.
4) Check your homework and quizzes for errors, and rework anything you got wrong.
5) Did I mention "practice"? :wink:
Eliz.
Need help on this one...
. . . . .Integrate: \(\displaystyle \large{\int{\,\frac{1}{(x^2\,-\,9)^2}\,}dx}\)
. . . . .
\(\displaystyle \large{\int{\,\frac{1}{(x\,-\,3)^2(x\,+\,3)^2}\,}dx}\)
. . . . .
\(\displaystyle \large{\int{\,\frac{A}{(x\,+\,3)}\,}}+{{\,\frac{B}{(x\,+\,3)^2}\,}}+
{{\,\frac{C}{(x\,-\,3)}\,}}+{{\,\frac{C}{(x\,-\,3)^2}\,}}\)
. . . . .
\(\displaystyle {{{A}{(x\,-\,3)^2(x\,+\,3)}\,}}+{{{B}{(x\,-\,3)^2}\,}}+{{{C}{(x\,+\,3)^2(x\,-\,3)}\,}}
+{{{D}{(x\,+\,3)^2}\,}}\)
im I on the right track?
. . . . .Integrate: \(\displaystyle \large{\int{\,\frac{1}{(x^2\,-\,9)^2}\,}dx}\)
. . . . .
\(\displaystyle \large{\int{\,\frac{1}{(x\,-\,3)^2(x\,+\,3)^2}\,}dx}\)
. . . . .
\(\displaystyle \large{\int{\,\frac{A}{(x\,+\,3)}\,}}+{{\,\frac{B}{(x\,+\,3)^2}\,}}+
{{\,\frac{C}{(x\,-\,3)}\,}}+{{\,\frac{C}{(x\,-\,3)^2}\,}}\)
. . . . .
\(\displaystyle {{{A}{(x\,-\,3)^2(x\,+\,3)}\,}}+{{{B}{(x\,-\,3)^2}\,}}+{{{C}{(x\,+\,3)^2(x\,-\,3)}\,}}
+{{{D}{(x\,+\,3)^2}\,}}\)
im I on the right track?
- Joined
- Apr 12, 2005
- Messages
- 11,339
If all else fails, it never hurts to over define things. You'll just get a few parameters that come out zero, but that isn't the case on this one. You'll need them all.
\(\displaystyle \L \frac{A}{x+3}+\frac{B}{(x+3)^{2}}+\frac{C}{x-3}+\frac{D}{(x-3)^{2}}\)
Note: Some don't like that version and choose to note that the denominators are inherently quadratic and that should be recognized.
\(\displaystyle \L \frac{Ax+B}{(x+3)^{2}}+\frac{Cx+D}{(x-3)^{2}}\)
Same thing, really.
Choose the way that makes the most sense to you.
\(\displaystyle \L \frac{A}{x+3}+\frac{B}{(x+3)^{2}}+\frac{C}{x-3}+\frac{D}{(x-3)^{2}}\)
Note: Some don't like that version and choose to note that the denominators are inherently quadratic and that should be recognized.
\(\displaystyle \L \frac{Ax+B}{(x+3)^{2}}+\frac{Cx+D}{(x-3)^{2}}\)
Same thing, really.
Choose the way that makes the most sense to you.