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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluvate the following intregals:
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\ \text{dx}$
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Answer

Let $\text{I}=\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\ \text{dx}$
$\text{x}+2=\text{A}\frac{\text{d}}{\text{dx}}[\text{x}^2+2\text{x}+3]+\text{B}$
$\Rightarrow\text{x}+2=2\text{Ax}+2\text{A}+\text{B}$
Comparing the coefficient, we have,
$2\text{A}=1\text{ and }2\text{A}+\text{B}=2$
$\Rightarrow\text{A}=\frac{1}{2}$
Substituting the value of A in 2A + B = 2, we have,
$2\times\frac{1}{2}+\text{B}=2$
$\Rightarrow1+\text{B}=2$
$\Rightarrow\text{B}=2-1$
$\Rightarrow\text{B}=1$
Thus we have,
$\text{x}+2=\frac{1}{2}[2\text{x}+2]+1$
hence,
$\text{I}=\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\text{dx}$
$=\int\frac{\big[\frac{1}{2}[2\text{x}+2]+1\big]}{\sqrt{\text{x}^2+2\text{x}+3}}\text{dx}$
$=\int\frac{\big[\frac{1}{2}[2\text{x}+2]\big]}{\sqrt{\text{x}^2+2\text{x}+3}}\text{dx}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+3}}$
$=\frac{1}{2}\int\frac{[2\text{x}+2]}{\sqrt{\text{x}^2+2\text{x}+3}}\text{dx}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+3}}$
Substituting t = x2 + 2x + 3 and dt = 2x + 2 in the first intrgrand, we have,
$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+3}}$
$=\frac{1}{2}\times2\sqrt{\text{t}}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+1+2}}+\text{C}$
$=\sqrt{\text{t}}+\int\frac{\text{dx}}{\sqrt{(\text{x}+1)^2+(\sqrt{2}})^2}+\text{C}$
$\text{I}=\sqrt{\text{x}^2+2\text{x}+3}+\log\big[|\text{x}+1|+\sqrt{(\text{x}+1)^2+(\sqrt{2}}^2\Big]+\text{C}$
$\text{I}=\sqrt{\text{x}^2+2\text{x}+3}+\log\Big[|\text{x}+1|+\sqrt{\text{x}^2+2\text{x}+3}\Big]+\text{C}$

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