Gujarat 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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