Evalute the following integrals: 1 (x+1)^2(x^2+1) dx - Vidyadip

Question

Evalute the following integrals:
$\int\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\ \text{dx}$

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Answer

We have,
$\text{I}=\int\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\ \text{dx}$
Let $\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{(\text{x}+1)^2}+\frac{\text{Cx}+\text{D}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}(\text{x}+1)(\text{x}^2+1)+\text{B}(\text{x}^2+1)(\text{Cx}+\text{D})(\text{x}+)^2}{(\text{x}+1)^2(\text{x}^2+1)}$
$\Rightarrow1=\text{A}(\text{x}^3+\text{x}+\text{x}^2+1)+\text{B}(\text{x}^2+1)+(\text{Cx}+\text{D})(\text{x}^2+2\text{x}+1)$
$\Rightarrow1=\text{A}(\text{x}^3+\text{x}^2+\text{x}+1)+\text{B}(\text{x}^2+1)+\text{Cx}^3+2\text{Cx}^2+\text{Cx}+\text{Dx}^2+2\text{Dx}+\text{D}$
$\Rightarrow1=(\text{A}+\text{C})\text{x}^3+(\text{A}+\text{B}+2\text{C}+\text{D})\text{x}^2+(\text{A}+\text{C}+2\text{D})\text{x}+\text{A}+\text{B}+\text{D}$
Equating coefficients of like terms
A + C = 0 ...(1)
A + B + 2C + D = 0 ...(2)
A + C + 2D = 0 ...(3)
A + B + D = 1 ...(4)
$\text{A}=\frac{1}{2},\text{B}=\frac{1}{2},\text{C}=-\frac{1}2{}$ and $\text{D}=0$
$\therefore\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{1}{2(\text{x}+1)}+\frac{1}{2(\text{x}+1)^2}-\frac{1}{2}\times\frac{\text{x}}{\text{x}^2+1}$
$\Rightarrow\int\frac{\text{dx}}{(\text{x}+1)^2(\text{x}^+1)}=\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{2}\int\frac{\text{dx}}{(\text{x}+1)^2}-\frac{1}{2}\int\frac{\text{x dx}}{\text{x}^2+1}$
Putting $\text{x}^2+1=\text{dt}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{2}\int\frac{\text{dx}}{(\text{x}+1)^2}-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{2}\log|\text{x}+1|-\frac{1}{2(\text{x}+1)}-\frac{1}{4}\log|\text{t}|+\text{C}'$
$=\frac{1}{2}\log|\text{x}+1|-\frac{1}{2(\text{x}+1)}-\frac{1}{4}\log|\text{x}^2+1|+\text{C}'$

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