I can't figure out how to do this problem.
INT.[1 / {x(x^4 +1)}]dx
Mutiply both numerator and denominator by x^3,
= INT.[x^3 / {(x^4)(x^4 +1)}] dx
= INT.[1 / {(x^4)(x^4 +1)}] (x^3 dx) ------------(i)
Let u = x^4
So, du = (4x^3)dx
Then (i) becomes
= INT.[1 / (u*(u+1))]*(1/4)du
= (1/4)INT.[1 / u(u+1)]du ------------------(ii)
By partial fraction:
1 / u(u+1) = A/u +B/(u+1)
Multiply both sides by u(u+1),
1 = A(u+1) +B*u ------------------(a)
When u = -1, in (a),
1 = 0 +B(-1)
So, B = -1
When u = 0, in (a),
1 = A(0+1) +0
So, A = 1
Hence, (ii) becomes
= (1/4)INT.[1/u -1/(u+1)]du
= (1/4)ln(u) -(1/4)ln(u+1) +C ---------(iii)
Return to u = x^4,
= (1/4)ln(x^4) -(1/4)ln(x^4 +1) +C
= ln[(x^4)^(1/4)] -(1/4)ln(x^4 +1) +C
= ln(x) -(1/4)ln(x^4 +1) +C --------------answer.