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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals3 Marks
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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