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Integrate the rational functions
$\frac{1}{x\left(x^n+1\right)}$ [Hint: multiply numerator and denominator by $x^{n-1}$ and put $x^n=t$ ]
$\frac{1}{x\left(x^n+1\right)}$ [Hint: multiply numerator and denominator by $x^{n-1}$ and put $x^n=t$ ]
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Verified Answer
$\frac{x^{n-1}}{x \cdot x^{n-1}\left(x^n+1\right)}=\frac{x^{n-1}}{x^n\left(x^n+1\right)}$
Put $x^n=t$ so that $n x^{n-1} d x=d t$
$\therefore \int \frac{d x}{x\left(x^n+1\right)}=\frac{1}{n} \int \frac{d t}{t(t+1)} \quad \ldots(i)$
let $\frac{1}{t(t+1)} \equiv \frac{A}{t}+\frac{B}{t+1}$
$\Rightarrow 1 \equiv \mathrm{A}(\mathrm{t}+1)+\mathrm{Bt} \quad \ldots(ii)$
Put $\mathrm{t}=0,-1$ in (i), we get : $\Rightarrow \mathrm{A}=1 \& \mathrm{~B}=-1$
$\therefore \int \frac{d x}{x\left(x^n+1\right)}=\frac{1}{n} \int \cdot\left(\frac{1}{t}-\frac{1}{t+1}\right) d t$
$=\frac{1}{n}[\log |t|-\log |t+1|]+c=\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+c$
Put $x^n=t$ so that $n x^{n-1} d x=d t$
$\therefore \int \frac{d x}{x\left(x^n+1\right)}=\frac{1}{n} \int \frac{d t}{t(t+1)} \quad \ldots(i)$
let $\frac{1}{t(t+1)} \equiv \frac{A}{t}+\frac{B}{t+1}$
$\Rightarrow 1 \equiv \mathrm{A}(\mathrm{t}+1)+\mathrm{Bt} \quad \ldots(ii)$
Put $\mathrm{t}=0,-1$ in (i), we get : $\Rightarrow \mathrm{A}=1 \& \mathrm{~B}=-1$
$\therefore \int \frac{d x}{x\left(x^n+1\right)}=\frac{1}{n} \int \cdot\left(\frac{1}{t}-\frac{1}{t+1}\right) d t$
$=\frac{1}{n}[\log |t|-\log |t+1|]+c=\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+c$
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