Integrals Indefinite

3.1/1+x³=1/(x+1)(x²-x+1)=A/(x+1)+(Bx+c)/(x²-x+1)

Fr 1) Have already tried by partial fraction, better approach ki talaash mein.

@rahul... thnx buddy, but i don't like this A,B,C method at all, bahut lamba pakau sort of it is ! [144]

5)

http://targetiit.com/iit-jee-forum/posts/integrate-2-13927.html

3.
will this help ?
1+x^3=(1+x)(1-x+x^2)
and then use partial fractions ;)

Thnx fr the link Che.

one more way is there it was given by kaymant sir lomg back
http://targetiit.com/iit-jee-forum/posts/for-all-u-iitians-out-there-even-my-fiitjee-sir-co-10832.html
but thats wwat i have applied

∫1/(1+tanx)=∫cosx/(cos+sinx)

=1/√2∫cosxsin(x+pi/4)dx

sybtitute
x+pi/4=t

1/√2∫cos(t-pi/4)sintdt

1/√2∫cos(t)cos pi/4-sintsinpi/4sintdt

(1/2)∫cot(t)-(1/√2)∫sinpi/4dt