I can't figure out how to do this problem.
$\displaystyle
\int \frac {1}{x(x^4+1)}
dx
$
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.
Note that
$\displaystyle \frac{1}
{{x\left( {x^4 + 1} \right)}} = \frac{{\left( {x^4 + 1} \right) - x^4 }}
{{x\left( {x^4 + 1} \right)}} = \frac{1}
{x} - \frac{{x^3 }}
{{x^4 + 1}}$
Integrating
$\displaystyle \int {\frac{1}
{{x\left( {x^4 + 1} \right)}}\,dx} = \int {\frac{1}
{x}\,dx} - \frac{1}
{4}\int {\frac{{\left( {x^4 + 1} \right)'}}
{{x^4 + 1}}\,dx} = \ln \left| x \right| - \frac{1}
{4}\ln \left( {x^4 + 1} \right) + k$
From there (if you desire), you may do some of make-up with the natural log.