4I = ∫dx/[x(1/x4 + 1)]
Put 1/x4 + 1 = t4
=> dx/x = - dt [t3/(1-t)]
I = ∫[t3/(1-t)] dt
Put 1-t= p
=> - dt = dp
I = -∫(1-p)3/p dp
= - ln p + p3/3 + 3p - 3p2/2
Now substitute x in terms of p
3 Answers
4
3actually indef not all that important ... hardly 3 marks ull get frm it in jee ..... so dont take tension....
39never underestimate any lesson , who knows h e might just put up a matrix on it
