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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
$\int\frac{\text{x}^2+3\text{x}-1}{(\text{x}+1)^2}\text{dx}$
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Answer

$\text{Let I}=\int\frac{\text{x}^2+3\text{x}-1}{(\text{x}+1)^2}\text{dx. Then}$$\text{I}=\int\frac{\text{x}^2+\text{x}+2\text{x}-1}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}(\text{x}+1)+2\text{x}-1}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}(\text{x}+1)}{(\text{x}+1)^2}\text{dx}+\int\frac{2\text{x}-1}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}}{\text{x}+1}\text{dx}+\int\frac{\sqrt{2\text{x}+2-2-1}}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}+1-1}{\text{x}+1}\text{dx}+\int\frac{2(\text{x}+1)-3}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}+1}{\text{x}+1}\text{dx}-\int\frac{1}{\text{x}+1}\text{dx}+\int\frac{2(\text{x}+1)}{(\text{x}+1)^2}\text{dx}-3\int\frac{1}{(\text{x}+1)^2}\text{dx}$
$=\int\text{dx}-\int\frac{1}{\text{x}+1}\text{dx}+2\int\frac{1}{\text{x}+1}\text{dx}-3\int(\text{x}+1)^{-2}\text{dx}$
$=\text{x}-\log|\text{x}+1|+2\log|\text{x}+1|+\frac{3}{\text{x}+1}+\text{C}$
$=\text{x}+\log|\text{x}+1|+\frac{3}{\text{x}+1}+\text{C}$
$\therefore\text{I}=\text{x}+\log|\text{x}+1|+\frac{3}{\text{x}+1}+\text{C}$

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