IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

chinmay_saxena01

integration of x^3+x/x^4-9

here we can seperate this as ,,,,x^3/x^4-9 + x/x^4-9

but i want to know can we ssolve this by supposing num = kd/dx(den.) + q

and plz tell me when shud we apply this method....bcoz most of question r possible thru this one or others ,,,,and i can not get the key idea quickly to solve this....

studyid

Probably in this case .... doing the integral by supposing num = kd/dx(den.) + q ...... wud be a little difficult .........

a better option wud be to carry forward by your first method ......

the first integral is easy to evaluate ...... while in the other integral put
(x)^(2) = t and evaluate :)

chinmay_saxena01

ok i agree with u but when shud i apply 2nd method.....

chinmay_saxena01

even i applied this method in my question but i got the different answer.....

chinmay_saxena01

yup the answer is
1/4log|x^4-9| +1/12log|x^2-3/x^2+3| +c

chinmay_saxena01

shud i explain it also or not

mukundmadhav

These are two different terms right.. So where does the question of expressing numerator as d/dx of denominator come?

chinmay_saxena01

see here is the sol.
first seperate this as......
integrationx^3/x^4-9 +integrationx/x^4-9
then it is as I = p+q
so for p x^4=t then 4x^3dx =dt
and for 2nd one write den. as (x^2)^2-(3)^2 and put x^2 = t u will get answer on further integrating.....

chinmay_saxena01

hey madhav u didn't read my question care fully first go thru the question ....

mukundmadhav

x^3 + x is the numerator?
Put brackets yaar.. Kaise samjhega

mukundmadhav

Ya so your method is alright.. What's the problem?

chinmay_saxena01

hey i want to say ,,,,
take x as common from num. ok......
now write den as (x^2)^2-(3)^2 and let x^2 = t so we get 2xdx=dt and now we can apply that method which i mentioned in my problem,,,,,

mukundmadhav

yeah you can do that too..

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