Question
Evaluate the following intregals:
$\int\frac{\text{x}^4}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
$\int\frac{\text{x}^4}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
✓
Answer
we have
$\text{I}=\int\frac{\text{x}^4\text{dx}}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
$=\int\Big[\frac{\text{x}^4-1+1}{(\text{x}-1)(\text{x}^2+1)}\Big]\text{dx}$
$=\int\frac{(\text{x}^4-1)\text{dx}}{(\text{x}-1)(\text{x}^2+1)}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int\frac{(\text{x}^2-1)(\text{x}^2+1)\text{ dx}}{(\text{x}-1)(\text{x}^2+1)}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int\frac{(\text{x}-1)(\text{x}-1)\text{ dx}}{(\text{x}-1)}+\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int(\text{x}+1)\ \text{dx}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}\ ...(1)$
Let $\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}-1)}{(\text{x}-1)(\text{x}^2+1)}$
$\Rightarrow 1 = Ax^2 + A + Bx^2 - Bx + Cx - C$
$\Rightarrow 1 = (A + B)x^2 +(C - B)x + A - C$
Equating coefficient of like terms
A + B = 0 ...(1)
C - B = 0 ...(2)
A - C = 1 ...(3)
Solving (1), (2), (3) we get
$\text{B}=-\frac{1}{2},\text{A}=\frac{1}{2},\text{C}=-\frac{1}{2}$
$\therefore\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{1}{2(\text{x}-1)}+\frac{-\frac{\text{x}}{2}-\frac{1}{2}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{1}{2(\text{x}-1)}-\frac{1}{2}\Big(\frac{\text{x}}{\text{x}^2+1}\Big)-\frac{1}{2(\text{x}^2+1)}\ ...(2)$
From (1) and (2)
$\text{I}=\int(\text{x}+1)\ \text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{x dx}}{\text{x}^2+1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
Putting $\text{x}^2+1=\text{t}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int(\text{x}+1)\ \text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$=\frac{\text{x}^2}{2}+\text{x}+\frac{1}{2}\log|\text{x}-1|-\frac{1}{4}\log|\text{t}|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
$\text{I}=\int\frac{\text{x}^4\text{dx}}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
$=\int\Big[\frac{\text{x}^4-1+1}{(\text{x}-1)(\text{x}^2+1)}\Big]\text{dx}$
$=\int\frac{(\text{x}^4-1)\text{dx}}{(\text{x}-1)(\text{x}^2+1)}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int\frac{(\text{x}^2-1)(\text{x}^2+1)\text{ dx}}{(\text{x}-1)(\text{x}^2+1)}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int\frac{(\text{x}-1)(\text{x}-1)\text{ dx}}{(\text{x}-1)}+\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int(\text{x}+1)\ \text{dx}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}\ ...(1)$
Let $\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}-1)}{(\text{x}-1)(\text{x}^2+1)}$
$\Rightarrow 1 = Ax^2 + A + Bx^2 - Bx + Cx - C$
$\Rightarrow 1 = (A + B)x^2 +(C - B)x + A - C$
Equating coefficient of like terms
A + B = 0 ...(1)
C - B = 0 ...(2)
A - C = 1 ...(3)
Solving (1), (2), (3) we get
$\text{B}=-\frac{1}{2},\text{A}=\frac{1}{2},\text{C}=-\frac{1}{2}$
$\therefore\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{1}{2(\text{x}-1)}+\frac{-\frac{\text{x}}{2}-\frac{1}{2}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{1}{2(\text{x}-1)}-\frac{1}{2}\Big(\frac{\text{x}}{\text{x}^2+1}\Big)-\frac{1}{2(\text{x}^2+1)}\ ...(2)$
From (1) and (2)
$\text{I}=\int(\text{x}+1)\ \text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{x dx}}{\text{x}^2+1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
Putting $\text{x}^2+1=\text{t}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int(\text{x}+1)\ \text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$=\frac{\text{x}^2}{2}+\text{x}+\frac{1}{2}\log|\text{x}-1|-\frac{1}{4}\log|\text{t}|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
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