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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the follwing intregals:
$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
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Answer

$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
Let $\text{x} ^2=\text{y}$
$\Rightarrow2\text{x dx}=\text{dy}$
$\Rightarrow\text{dx}=\frac{\text{dy}}{2\text{x}}$
$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
$=\int\frac{\text{dy}}{(\text{y}+1)(\text{y}+2)^2}$
Let $\frac{1}{(\text{y}+1)(\text{y}+2)^2}=\frac{\text{A}}{\text{y}+1}+\frac{\text{B}}{\text{y}+2}+\frac{\text{C}}{(\text{y}+2)^2}\ ...(1)$
$\Rightarrow1=\text{A}(\text{y}+2)^2+\text{B}(\text{y}+1)(\text{y}+2)+\text{C}(\text{y}+1)\ ...(2)$
Putting y $= -2 in (2)$
$1 = C (-2 + 1)$
$\Rightarrow C = -1$
Putting $y = -1 in (2)$
$1 = A (-1 + 2)^2$
$\Rightarrow 1 = A (1)$
$\Rightarrow A = 1$
Putting y = 0 in $(2)$
$1 = 4A + B(2) + C$
$\Rightarrow 1 = 4 + 2B - 1$
$\Rightarrow -2 = 2B$
$\Rightarrow B = -1$
Substituting the values of A, B and C in (1)
$\frac{1}{(\text{y}+1)(\text{y}+2)^2}=\frac{1}{\text{y}+1}-\frac{1}{\text{y}+2}-\frac{1}{(\text{y}+2)^2}$
$\Rightarrow\int\frac{\text{dy}}{(\text{y}+1)(\text{y}+2)^2}=\int\frac{\text{dy}}{\text{y}+1}-\int\frac{\text{dy}}{\text{y}+2}-\int\frac{\text{dy}}{(\text{y}+2)^2}$
$=\log|\text{y}+1|-\log|\text{y}+2|+\frac{1}{\text{y}+2}+\text{C}$
Hence, $\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}=\log|\text{x}^2+1|-\log|\text{x}^2+2|+\frac{1}{\text{x}^2+2}+\text{C}$

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