Evaluate the following integrals: dx (x^2+1)(x^2+4)

Question

Evaluate the following integrals:
$\int\frac{\text{dx}}{(\text{x}^2+1)(\text{x}^2+4)}$

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Answer

Let $\frac{1}{(\text{x}^2+1)(\text{x}^2+4)}=\frac{\text{Ax}+\text{B}}{(\text{x}^2+1)}+\frac{\text{Cx}+\text{D}}{\text{x}^2+4}$
$\Rightarrow1=(\text{Ax}+\text{B})(\text{x}^2+4)+(\text{Cx}+\text{D})(\text{x}^2+1)$
$=(\text{A}+\text{C})\text{x}^3+(\text{B}+\text{D})\text{x}^2+(4\text{A}+\text{C})\text{x}+4\text{B}+\text{D}$
Equating similar terms, we get,
A + C = 0, B + D = 0, 4A + C = 0, 4B + D = 1
Solving, we get, $\text{A}=0,\text{B}=\frac{1}{3},\text{C}=0,\text{D}=-\frac{1}{3}$
Thus,
$\text{I}=\int\frac{\frac{1}{3}\text{dx}}{(\text{x}^2+1)}-\int\frac{\frac{1}{3}\text{dx}}{(\text{x}^2+4)}$
$=\frac{1}{3}\tan^{-1}\text{x}-\frac{1}{6}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$ $\big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}\big]$
$\therefore\text{I}=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$

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