# Integral Problem

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• Oct 14th 2007, 08:57 PM
adnaP
Integral Problem
I can't figure out how to do this problem.
$\displaystyle \int \frac {1}{x(x^4+1)} dx$
• Oct 14th 2007, 11:48 PM
ticbol
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.
• Oct 15th 2007, 09:18 AM
Krizalid
Quote:

Originally Posted by adnaP
$\displaystyle \int \frac {1}{x(x^4+1)} dx$

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.