Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Find $\int \frac{x^{4} d x}{(x-1)\left(x^{2}+1\right)}$
✓
Answer
We have $\frac{x^{4}}{(x-1)\left(x^{2}+1\right)}=(x+1)+\frac{1}{x^{3}-x^{2}+x-1}$
Using Partitial Fraction,
$= (x+1)+\frac{1}{(x-1)\left(x^{2}+1\right)} ......(i)$
Now express $\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{\mathrm{A}}{(x-1)}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^{2}+1\right)} .....(ii)$
So $1 = A (x^2 + 1) + (Bx + C) (x - 1)$
$= (A + B) x^2 + (C - B) x + A - C$
Equating coefficients on both sides, we get $A + B = 0, C - B = 0$ and $A - C = 1,$ which gives
$A=\frac{1}{2}, B=C=-\frac{1}{2}$
Substituting values of $A, B$ and $C$ in $(ii),$ we get
$\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)} .....(iii)$
Again, substituting $(iii)$ in $(i),$ we have
$\frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)} = (x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}$
Therefore
$\int \frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)} d x = \frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+C$
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