Evalute the following integrals: 1 1+x+x^2+x^3 dx - Vidyadip

Question

Evalute the following integrals:
$\int\frac{1}{1+\text{x}+\text{x}^2+\text{x}^3}\text{ dx}$

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Answer

We have,
$\text{I}=\int\frac{\text{dx}}{1+\text{x}+\text{x}^2+\text{x}^3}$
$=\int\frac{\text{dx}}{(\text{x}+1)(\text{x}^2+1)}$
$=\int\frac{\text{dx}}{(\text{x}+1)(\text{x}^2+1)}$
Let $\frac{1}{(\text{x}+1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}+1)(\text{x}^2+1)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+1)}{(\text{x}+1)(\text{x}^2+1)}$
$\Rightarrow1=\text{A}(\text{x}^2+1)+\text{Bx}^2+\text{Bx}+\text{Cx}+\text{C}$
$\Rightarrow1=(\text{A}+\text{B})\text{x}^2+(\text{B}+\text{C})\text{x}+(\text{A}+\text{C})$
Equating coefficient of like terms
A + B = 0 ...(1)
B + C = 0 ...(2)
A + C = 1 ...(3)
Solving (1), (2) and (3), we get
$\text{A}=\frac{1}{2}$
$\text{B}=-\frac{1}{2}$
$\text{C}=\frac{1}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{2}\int\Big(\frac{-\text{x}+1}{\text{x}^2+1}\Big)\text{dx}$
$=\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}-\frac{1}{2}\int\frac{\text{x dx}}{\text{x}^2+1}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1^2}$
Let $\text{x}^2+1=\text{dt}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1^2}$
$=\frac{1}{2}\log|\text{x}+1|-\frac{1}{4}\log|\text{t}|+\frac{1}{2}\tan^{-1}(\text{x})+\text{C}$
$=\frac{1}{2}\log|\text{x}+1|-\frac{1}{4}\log|\text{x}^2+1|+\frac{1}{2}\tan^{-1}(\text{x})+\text{C}$

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