# Integrate the rational functions$\frac{1}{x(x^n+1)}$ $Hint:multiply\;numerator\;and\;denominator\;by\;x^{n-1}and\;put\;x^n=t$

## 1 Answer

Toolbox:
• $(i)\;$Form of the ration function$\frac{px+q}{(x+a)(x+b)}$
• $\;$Form of the partial function$\frac{A}{(x+a)}+\frac{B}{(x+b)}$
• $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
Given:$I=\int\frac{1}{x(x^n+1}dx.$

Multiply the numerator and denominator by $x^n-1$.

then,$I=\int\frac{x^{n-1}}{x(x^n+1)(x^{n-1}}dx.$

But $x'(x^{n-1})=x^n.$

Therefore $I=\int\frac{x^{n-1}}{x^n(x^n+1)}dx.$

Now put $x^n=t,$ then $nx^{n-1}dx=dt.$

$\Rightarrow x^{n-1}dx=\frac{dt}{n}.$

Substituting this we get,

$I=\int\frac{dt/n}{t(t+1)}=\frac{1}{n}\int\frac{dt}{t(t+1)}$.

Resolving into partial fraction

Let $\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1}.$

$\Rightarrow 1=A(t+1)+Bt.$

Now equating the coefficient of t we get,

0=A+B-------(1)

Now equating the constant term,

1=A,Hence A=1.

Substituting for A in equ (1)

1+B=0$\Rightarrow B=-1.$Hence A=1 and B=-1.$Now substituting for A and B,we get$\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}.$Therefore$I=\frac{1}{n}\begin{bmatrix}\int\frac{1}{t}dt-\int\frac{1}{t+1}dt\end{bmatrix}$On integrating we get,$\frac{1}{n}log|t|-\frac{1}{n}log|t+1|+C.$Substituting for t we get,$\;\;\;=\frac{1}{n}log|x^n|-\frac{1}{n}log|x^n+1|+C.log a-log b=log\mid\frac{a}{b}\mid$,similarly$\;\;\;=\frac{1}{n}log\mid\frac{x^n}{x^n+1}\mid+c.\$

answered Feb 7, 2013

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