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Mathematics 16 Online
OpenStudy (cherrilyn):

evaluate the integral

OpenStudy (cherrilyn):

\[\int\limits_{?}^{?}100xdx/(x-3)(x ^{2}+1)^{2}\] I know I change it to A/x-1 + B(x-1)^2+C(x-1)^3 then multiply it by (x-1)^3 then what

OpenStudy (anonymous):

are you sure?

OpenStudy (cherrilyn):

oops I mean A/x-3 + Bx+C/x^2+1 + Dx+E/(x^2+1)^2

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

alright, good , now start taking small values for x ^_^ to find A, B, C and D

OpenStudy (anonymous):

wait, first you have to multiply the denominator on all parts

OpenStudy (anonymous):

Take x=3. You will find A directly.

OpenStudy (anonymous):

A=3

OpenStudy (anonymous):

so, you'll get :\[100x = A(x+1)^2 + Bx + C(x-3)(x^2+1) + Dx + E (x-3)\]

OpenStudy (cherrilyn):

ok A=3 :)

OpenStudy (cherrilyn):

wait no

OpenStudy (anonymous):

now you can take values for x to find all the alphabets ^_^

OpenStudy (anonymous):

hmm, what?

OpenStudy (cherrilyn):

when you distribute in..for example.. Bx+C(x-3)(x^2+1) ... do you distribute the two things in the parenthesis to the Bx AND the C? or just the C

OpenStudy (cherrilyn):

and Distribute (x-3) to the Dx AND the E?

OpenStudy (anonymous):

hmm, I think it's the whole thing

OpenStudy (anonymous):

yes yes, I'm sure

OpenStudy (cherrilyn):

okay so A is indeed 3

OpenStudy (anonymous):

yes it is ^_^

OpenStudy (anonymous):

The better way is to open all brackets, and add like terms together.

OpenStudy (anonymous):

now take other values for x, such as -1 1 2 -2 ~

OpenStudy (anonymous):

B=0

OpenStudy (anonymous):

Oh wait!!

OpenStudy (cherrilyn):

blaah do I have to test all of those numbers? lol its kiind of hard to predict the right values just by looking at it

OpenStudy (anonymous):

yes it's a must, you need to get all the values, A, B,C,D,and E

OpenStudy (cherrilyn):

I mean the numbers -1, 1, 2, and -2

OpenStudy (anonymous):

yes , you must lol, what other way is there to solve it?

OpenStudy (anonymous):

you must check with all values my dear :)

OpenStudy (anonymous):

B=-3

OpenStudy (anonymous):

C=9

OpenStudy (anonymous):

you checked the values? anwar? or shall I check after you lol

OpenStudy (anonymous):

D=18

OpenStudy (cherrilyn):

Okay If I plug in A=3 and x=-1..... I get -100=12+2B-2C+4D-4E. now what? Or am I doing this wrong

OpenStudy (anonymous):

no you're write, so now you have equation 1, you can use it later to either subtract or add from another equation ^_^

OpenStudy (anonymous):

right*

OpenStudy (anonymous):

alright I gtg, I'll be back later, bye ^_^

OpenStudy (cherrilyn):

okay thankss! Ttyl :)

OpenStudy (anonymous):

anwar will continue, most welcome :)

OpenStudy (anonymous):

I got all of them. Hopefully right!

OpenStudy (anonymous):

A=3, B=-3, C=9, D=24, E=-8

OpenStudy (anonymous):

Bye starica!!

OpenStudy (cherrilyn):

thanks anwar!

OpenStudy (anonymous):

You're welcome!! You may want to check them.

OpenStudy (anonymous):

OMG, Sorry there was a mistake.

OpenStudy (anonymous):

It should be: A=3, B=-3, C=-9, D=-30, E=10. Sorry for the confusion.

OpenStudy (cherrilyn):

what values for x did you use?

OpenStudy (anonymous):

Well.. I opened all the parenthesis, and add like terms together, and then solved the equations.

OpenStudy (anonymous):

For example if you have x^2(A+B)+x(C+2B)=5x^2+x, that means: A+BC=5 and C+2B=1.. I used this idea.

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