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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Integrate the function in Exercise:

$\frac{1}{\big(\text{x}^{2}+1\big)\big(\text{x}^{2}+4\big)}$

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Answer

$\therefore\frac{1}{\big(\text{x}^{2}+1\big)\big(\text{x}^{2}+4\big)}=\frac{\text{Ax}+\text{B}}{\big(\text{x}^{2}+1\big)}+\frac{\text{Cx}+\text{D}}{\big(\text{x}^{2}+4\big)}$
$\Rightarrow1=\big(\text{Ax}+\text{B}\big)\big(\text{x}^{2}+4\big)+\big(\text{Cx}+\text{D}\big)\big(\text{x}^{2}+1\big)$
$\Rightarrow1=\text{Ax}^{3}+4\text{Ax}+\text{Bx}^{2}+4\text{B}+\text{Cx}^{3}+\text{Cx}+\text{Dx}^{2}+\text{D}$
Equating the coefficients of $\text{x}^{3},\text{x}^{2}$ and constant term, we obtain
$\text{A}+\text{C}=0$
$\text{B}+\text{D}=0$
$4\text{A}+\text{C}=0$
$4\text{B}+\text{D}=1$
On solving these equations, we obtain
$\text{A}=0,\text{B}=\frac{1}{3},\text{C}=0,$and $\text{D}=-\frac{1}{3}$
From equation (1), we obtain
$\frac{1}{\big(\text{x}^{2}+1\big)\big(\text{x}^{2}+4\big)}=\frac{1}{3\big(\text{x}^{2}+1\big)}-\frac{1}{3\big(\text{x}^{2}+4\big)}$
$\int\frac{1}{\big(\text{x}^{2}+4\big)\big(\text{x}^{2}+4\big)}\text{dx}=\frac{1}{3}\int\frac{1}{\text{x}^{2}+1}\text{dx}-\frac{1}{3}\int\frac{1}{\text{x}^{2}+4}\text{dx}$
 $=\frac{1}{3}\tan^{-1}\text{x}-\frac{1}{3}.\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\text{C}$
$=\frac{1}{3}\tan^{-1}\text{x}-\frac{1}{6}\tan^{-1}\frac{\text{x}}{2}+\text{C}$

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