IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

res_saj

For me, resolving into partial fractions will take some time... coz our teachers at school taught us that you can either
1. compare coeff. and then find the values of 'A', 'B' etc.
2. substitute values (so that one term becomes zero) and then find A, do the same thing for B etc.

But when I'm going through this Dinesh Objective Mathematics ,they just do it in one step .. is there any other method for resolving into partial fractions?

Thanx
Reshma

sushu

hey reshma

give me 1 eg. i'll solve it n explain u how to do it.

bye.

ritu_007

hi! using the second method the prob.can be solved in a single step.

res_saj

ok.. for example.. this :

z²/(1+z²)(1-z²)
Thnx,
Reshma

sushu

hi reshma

iam here 4 u again.let z=x for our convienience.

now x²/(1+x²)(1-x²)=ax+b/(1+x²) +c/(1+x)+d/(1-x).................(1)

since 1-x²=(1-x)(1+x)

now comparing numerators we get x²=ax+b(1+x)(1-x)+c(1+x²)(1-x)+d(1+x²)(1+x)

put x=1 we get 1=d(2)(2) so d=1/4

put x=-1 we get 1=c(2)(2) so c=1/4

now comparing x³coeffs we get 0=-a-c+d so a=0

now compare x² coeffs we get 1=-b+c+d. so b=1/2

sub this in eq 1. u'll get d ans.

if it is wrong pls tell me.if it is rite pls rate me.

hope u understood it clearly.

sushu

hey reshma!

u there? reply me soon.

res_saj

I wz a little busy.. sorry

The thing is... I know what u said - I learnt it for my boards which is on the coming Thursday.

But its too time consuming... i asked if ther is any method to do it in "one" step - coz in the Dinesh.. they just do it in one step.
Or shuld I do all this mentally - to get in one step??

thnx,
Reshma

bubunbhatt

Hi

a simple metod For doing it;

Without using P.F.

S:rep int.

1/2 S 2z^2/(1+z^2)(1-z^2)

=1/2 S (1+z^2)-(1-z^2)/(1+z^2)(1-z^2)

Now divide and int.

It will come in Some Integrable form

It takes much less time

Comeon Rate me

ruhi

hi reshma..........there is not a specific short method.........i mean it depends on de ques...........partial fraction is a general one............. a bit long........but u ll get de ans 4 sure..........now check mr. bubun soln........his method is a short one.......... so v ve 2 apply our brain......how can a ques b short...........n dis can b easy.....by practice............hope u got it wot i wanna say :)

phyana

there is an easy method to solve the partial fractions.
solution 1.
if the denominator is expressed in the split form
ie.......(x+2)(x+3)...
then express the each part as an reciprocal and finally divide all the trems with the difference of the constants.
similarly there are other ways also...........

kiran

HI

There is one shortcut which works for all cases when the denominator contains x in 1 degree and some cases when x is in degree 2.

Suppose you have a polynomial (2x+3) / (x-2) (x-1) then the partial fractions are 7 / (x-2) - 5 / (x-1) .

This is done by just making a factor of denominator 0 by putting some value to it , and using the same value for other rest of the polynomial and getting a final value.

In the example the denominator has factors (x-2) (x-1)

Put x =2 and get the value of rest of the polynomial. ie (2x+3)/(x-1) at x =2 gives 7 . Now put x =1 ( for factor x-1) and you get -5 .

Hence the partial fractions are 7 / (x-2) - 5 (x-1) .

Even I am searching for quick methods for denominator which contains factors of degree 2 and degree 3.

Cheers!

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