Integrate the rational function 1 x (x^ n +1 ) [Hint: multiply nu… - Vidyadip

Question

Integrate the rational function $\frac{1}{x\left(x^{n}+1\right)} [$Hint: multiply numerator and denominator by $x^{n-1}$ and put $x^n = t]$

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Answer

Given function is, $\frac{1}{x\left(x^{n}+1\right)}$
Multiplying numerator and denominator by $x^{n-1},$ we get,
$\frac{1}{x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n-1} x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)}$
Let $x^n = t$
$nx^{n-1}dx = dt$
Therefore,
$\int \frac{1}{x\left(x^{n}+1\right)} d x=\int \frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} d x=\frac{1}{n} \int \frac{1}{t(t+1)} d t$
Let $\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}$
$1 = A(1 + t) + Bt ...(i)$
Substituting $t = 0, -1$ in equation $(i),$ we get,
$A =1$ and $B = -1$
Thus,
$\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(1+t)}$
$\int \frac{1}{x\left(x^{n}+1\right)} d x=\frac{1}{n} \int\left\{\frac{1}{t}-\frac{1}{(1+t)}\right\} d t$
$= \frac{1}{n}[\log |t|-\log |t+1|]+C$
$= \frac{1}{n}\left[\log \left|x^{n}\right|-\log \left|x^{n}+1\right|\right]+C$
$= \frac{1}{n} \log \left|\frac{x^{n}}{x^{n}+1}\right|+C$

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