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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra 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
⇒ 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
⇒ 2 = -B
or B = -2 ...(5)
putting value of (B) in (1), we get
10 = 4 (-2) + 3D
$\frac{10+8}{3}=\text{D}$
⇒ 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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