Integrate the rational function 1 x (x^ n +1 ) [Hint: multiply numerator and denominator by xn-1 and put xn = t]

Given function is, 1 x (x^ n +1 ) Multiplying numerator and denominator by xn-1, we get, 1 x (x^ n +1 ) = x^ n-1 x^ n-1 x (x^ n +1 ) = x^ n-1 x^ n (x^ n +1 ) Let xn = t nxn-1dx = dt Therefore, 1 x (x^ n +1 ) d x= x^ n-1 x^ n (x^ n +1 ) d x=1 n 1 t(t+1) d t Let 1 t(t+1) =A t +B (t+1) 1 = A(1 + t) + Bt ...(i) Substituting t = 0, -1 in equation (i), we get, A =1 and B = -1 Thus, 1 t(t+1) =1 t -1 (1+t) 1 x (x^ n +1 ) d x=1 n 1 t -1 (1+t) d t = 1 n [ |t|- |t+1|]+C = 1 n [ |x^ n |- |x^ n +1 | ]+C = 1 n | x^ n x^ n +1 |+C

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

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Question

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