I can't figure out how to do this problem.

$\displaystyle

\int \frac {1}{x(x^4+1)}

dx

$

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- Oct 14th 2007, 08:57 PMadnaPIntegral 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 PMticbol
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 AMKrizalid
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.