IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

abhijeet_0201

(x^1/2 dx)/(x^1/2-x^1/3) plz tell the ans as soon as possible

abhijeet_0201

i am waiting for an answer

rik_mad

hi,



(x^1/2 dx)/(x^1/2-x^1/3)

take x^1/2 common outside from denominator and cancel the common term from the fraction u will get

dx/(1 - x^(-1/6))

now substitute x^(1/6) with t and then u will get

=6(t^6)dt/(t-1)

= write numerator as

(6(t^6) + 6 - 6)dt/(t-1)

now separate into

(6(t^6) - 6)dt/(t-1) + 6dt/(t-1)

now second term is easily done

for first term we know that for (x^6 - 1)/x-1 = x^5 + x^4 + x^3 ...+1

use it in first term and get the answer

YO !

karthik_abiram

hii,

sorry,some mistake i have done

abhijeet_0201

i cant understand the steps after seprating.plz explain elaborately

nishant

well abhijeet,what rik has said that the second term can be solved easily ..this is because integration of dt/t-1 gives log mod t-1..also ,we have 6 outside the integration sign,so the second term becomes 6 log mod( t-1)..for first term,he has taken 6 common frm numerator and has solved it by using the fromula of a^3-b^3 in the numerator..hope this helps ..cheers..

abhijeet_0201

i got it now.thanks a lot to both.

Ask & Discuss Questions with Community & Experts