Well this week we talked about partial fractions and B integration tables. So I guess that is what I will be covering in this blog.
Partial Fractions:
problem: S 1/x^2+5x+6
first: factor the bottom
1/(x+3)(x+2)
second: set up the fractions, start off with a over the first, set equal to problem
A/x+3 + B/x+2 = 1/x^2+5x+6
third: multiply to get denominates equal, which gives you
A(x+2) + B(x+3) = 1
fourth: find convenient values for x to solve for A and B
A(x+2) + B(x+3)
*choose -2 to plug into A b/c it will give you zero
A(-2+2) + B(-2+3)=1
A(0) + B(1) =1
B=1
*choose -3 for B /c it will give you zero
A(-3+2) + B(-3+ -3)=1
A(-1) + B(0)=1
A=-1
fifth: plug A and B back into the fractions A/x+3 + B/x+2
S -1/x+3 + 1/x+2
*1/x gives you a ln
=-ln(x+3) + ln(x+2) +C
*simplify
ln(x+2/x+3) +C
Note: if you were given S fx^2+20x+6/x(x+1)^2, your fractions would be, it is a rule
A/x + B/x+1 + C/(x+1)^2
B integration tables:
This is from A21 in the book. It has a listing of all kinds of integrals and what they equal. You basically have to figure out which integral applies to the problem. If you need a u, a, or something you have to figure that out from the problem. All there is left is to plug in and simplify. The only tricky part is make sure you pick the right integral to use.
My Question:
Can anyone tell me which integral I would use for these two:
S 1/squareroot x(1-cos squareroot x)
or
S x^7lnx
*for this one there is a u^n formula nd a 1/u formula? but what do I use?
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ReplyDeleteS(x^7)(lnx), you use by parts.
Your u would be ln(x), and your dv would be x^7.
You get (1/8)(x^8)(lnx) - S(1/8)(x^7) + C
I couldn't really comment on anything because I wasn't here for the integration tables...but I do know that what Ryan said above is correct. you just use by parts.
ReplyDeleteHowever, for the first problem you mentioned, i'm confused as to what it is. Is it:
S 1(all over)/(sqrt(x)*(1-cos(sqrtx))??