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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following integrals:
$\int\frac{(\text{x}^2+1)(\text{x}^2+4)}{(\text{x}^2+3)(\text{x}^2-5)}\ \text{dx}$
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Answer

$\text{I}=\int\frac{(\text{x}^2+1)(\text{x}^2+4)}{(\text{x}^2+3)(\text{x}^2-5)}\ \text{dx}$
Since,
$\frac{(\text{x}^2+1)(\text{x}^2+4)}{(\text{x}^2+3)(\text{x}^2-5)}=\frac{[(\text{x}^2+3)-][(\text{x}^2-5)+9]}{(\text{x}^2+3)(\text{x}^2-5)}$
$\Rightarrow\frac{(\text{x}^2+1)(\text{x}^2+4)}{(\text{x}^2+3)(\text{x}^2-5)}=\frac{(\text{x}^2+3)(\text{x}^2-5)+9(\text{x}^2+3)-2(\text{x}^2-5)-18}{(\text{x}^2+3)(\text{x}^2-5)}$
$\Rightarrow\frac{(\text{x}^2+1)(\text{x}^2+4)}{(\text{x}^2+3)(\text{x}^2-5)}=1+\frac{9}{(\text{x}^2-5)}-\frac{2}{(\text{x}^2+3)}-\frac{18}{(\text{x}^2+3)(\text{x}^2-5)}\ ...(1)$
Let $\text{I}_1=\int\frac{1}{(\text{x}^2+3)(\text{x}^2-5)}$ and $\text{x}^2=\text{y}$
$\Rightarrow\frac{1}{(\text{y}+3)(\text{y}-5)}=\frac{\text{A}}{(\text{y}+3)}+\frac{\text{B}}{(\text{y}-5 )}$
$=\frac{\text{A}(\text{y}-5)+\text{B}(\text{y}+3)}{(\text{y}+3)(\text{y}-5)}$
$\Rightarrow\frac{1}{(\text{y}+3)(\text{y}-5)}=\frac{(\text{A}+\text{B})\text{y}-(5\text{A}+3\text{B})}{(\text{y}+3)(\text{y}-5)}$
Compairing the coefficient, we get
A + B = 0 and 5A + 3B = -1
By solving the equations, we get
$\text{A}=-\frac{1}{8}$ and $\text{B}=\frac{1}{8}$
From (1) we get
$\text{I}=\int\Big[1+\frac{9}{(\text{x}^2-5)}-\frac{2}{(\text{x}^2+3)}-18\Big(\frac{-1}{8(\text{x}^2+3)}+\frac{1}{8(\text{x}^2-5)}\Big)\Big]\text{dx}$
$\Rightarrow\text{I}\int\Big[1+\frac{27}{4(\text{x}^2-5)}+\frac{1}{(\text{x}^2+3)}\Big]\text{dx}$
$\therefore\text{I}=\text{x}+\frac{27}{8\sqrt{5}}\ln\Big(\big|\frac{\text{x}-\sqrt{5}}{\text{x}+\sqrt{5}}\big|\Big)+\frac{1}{4\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{3}}\Big)+\text{C}$

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