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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals3 Marks
Question
Integrate the rational function $\frac{1}{x\left(x^{4}-1\right)}$
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Answer

Given the function, $\frac{1}{x\left(x^{4}+1\right)}$
Multiplying numerator and denominator by $x^3,$ we get,
$\frac{1}{x\left(x^{4}+1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}+1\right)}$
Therefore, $\int \frac{1}{x\left(x^{4}+1\right)} d x=\int \frac{x^{3}}{x^{4}\left(x^{4}+1\right)} d x$
Now, let $x^4 = t$
$4x^3dx = dt$
Thus, $\int \frac{1}{x\left(x^{4}+1\right)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)}$
Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$
$1 = A(t - 1) + Bt …(i)$
Substituting $t = 0$ and $1$ in $(i),$ we get
$A = -1$ and $B = 1$
Therefore $\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$
$\int \frac{1}{x\left(x^{4}+1\right)} d\ x$
$=\frac{1}{4} \int \left\{\frac{-1}{t}+\frac{1}{t-1}\right\} d\ t$
$= \frac{1}{4}[-\log |t|+\log |t-1|]+C$
$= \frac{1}{4} \log \left|\frac{t-1}{t}\right|+C$
$= \frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+C$

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