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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{(\text{x}^2+1)(\text{x}^2+2)}{(\text{x}^2+3)(\text{x}^2+4)}\ \text{dx}$
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Answer

$\text{f}(\text{x})=\frac{(\text{x}^2+1)(\text{x}^2+2)}{(\text{x}^2+3)(\text{x}^2+4)}$
Now,
$\frac{(\text{x}^2+1)(\text{x}^2+2)}{(\text{x}^2+3)(\text{x}^2+4)}$
$=\frac{\text{x}^4+3\text{x}^2+2}{\text{x}^4+7\text{x}^2+12}$
$=\frac{(\text{x}^4+7\text{x}^2+12)-4\text{x}^2-10}{\text{x}^4+7\text{x}^2+12}$
$=1-\frac{4\text{x}^2+10}{\text{x}^4+7\text{x}^2+12}$
Now,
$\frac{4\text{x}^2+10}{\text{x}^4+7\text{x}^2+12}=\frac{4\text{x}^2+10}{(\text{x}^2+3)(\text{x }^2+4)}$
let $\frac{4\text{x}^2+10}{(\text{x}^2+3)(\text{x}^2+4)}=\frac{\text{Ax}+\text{B}}{\text{x}^2+3}+\frac{\text{Cx}+\text{D}}{\text{x}^2+4}$
$\Rightarrow4\text{x}^2+10=(\text{Ax}+\text{B})(\text{x}^2+4)+(\text{Cx}+\text{D})(\text{x}^2+3)$
let $x = 0, $ we get
$10 = 4B + 3D ...(1)$
If $x = 1$, we get
$14 = 5 (A + B) + 4 (C + D) = 5A + 5B + 4C + 4D ...(2)$
If $x = -1$, we get
$14 = 5 (-A + B) + 4 (-C + D) = -5A + 5B + -4C + 4D ...(3)$
Applying $(2)$ and $(3),$ we get
$28 = 10B + 8D$
$\Rightarrow 14 = 5B + 4D ...(4)$
From $(1)$ we get,
$10 = 4B + 3D$
Multiplying equations $(4)$ by $3$ and $(1)$ by $4$ and substracting, we get
$42 - 40 = 15B - 16G$
$\Rightarrow 2 = -B$
or $B = -2 ...(5)$
putting value of $(B)$ in 4 we get
$10 = 4 (-2) + 3D$
$\frac{10+8}{3}=\text{D}$
$\Rightarrow D = 6$
Compairing coefficient of $x^3 $ in
$4x^2 + 10 = (Ax + B)(x^2+ 4) + (Cx + 4)(x^2+3), we get,$
$0 = A + C ...(8)$
Compairing coefficient of x, we get
$0 = 4A + 3C$
$\Rightarrow A = C = 0$
$\therefore\text{f}(\text{x})=1-\frac{(-2)}{\text{x}^2+3}-\frac{6}{\text{x}^2+4}$
$=1+\frac{2}{\text{x}^2+3}-\frac{6}{\text{x}^2+4}$
$\therefore\int\text{f}(\text{x})\text{dx}=\int1+\frac{2}{\text{x}^2+3}-\frac{6}{\text{x}^2+4}\ \text{dx}$
$=\text{x}+\frac{2}{\sqrt{3}}\tan^{-1}\times\frac{\text{x}}{\sqrt{3}}-3\tan^{-1}\frac{\text{x}}{2}+\text{C}$

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