Question
$\text{Evaluate:}\int\frac{\text{x + 2}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}\text{dx}$
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Answer
$\text{Let I}=\int\frac{\text{x + 2}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}\text{dx}$$=\frac{\text{1}}{\text{2}}\int\frac{\text{2x + 4}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}\text{dx}=\frac{\text{1}}{\text{2}}\int\frac{\text{(2x + 2) + 2 }}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}\text{dx}$
$=\frac{\text{1}}{\text{2}}\int\frac{\text{(2x + 2)}\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}+\frac{\text{1}}{\text{2}}\int\frac{\text{2dx }}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}$
$\text{I}=\frac{\text{1}}{\text{2}}\text{I}_{1}+\text{I}_{2}-------(i)$
$\text{Where }\text{I}_{1}=\int\frac{\text{(2x+2)}\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}\text{ and }\text{I}_{2}=\int\frac{\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}$ $\text{I}_{1}=\int\frac{\text{2x+2}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}\text{dx}$ $x^2 + 2x + 3 = z^2$
(2x + 2)dx = 2z dz $\Rightarrow$ $\text{I}_{1}=\int\frac{\text{2z dz}}{\text{z}}$$=\text{2}\int{dz}=2z=2\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\text{C}_{1}$
$\Rightarrow$ $\text{I}_{1}=2\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\text{C}_{1}$ Again $\text{I}_{2}=\int\frac{\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}=\int\frac{\text{dx}}{\sqrt{\text{(x+1)}^{2}+(\sqrt{\text{2}}})^{2}}$$=\log|\text{(x+1)}+\sqrt{\text{(x+1)}^{2}+(\sqrt{\text{2}})^{2}}|$
$=\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{C}_{2}$
Putting the value of $I_1$, and $I_2$ in $(i)$ we get$\text{I}=\text{2}\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{(C}_{1}+\text{C}_{2})$
$=\text{2}\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{C.}$
$=\frac{\text{1}}{\text{2}}\int\frac{\text{(2x + 2)}\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}+\frac{\text{1}}{\text{2}}\int\frac{\text{2dx }}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}$
$\text{I}=\frac{\text{1}}{\text{2}}\text{I}_{1}+\text{I}_{2}-------(i)$
$\text{Where }\text{I}_{1}=\int\frac{\text{(2x+2)}\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}\text{ and }\text{I}_{2}=\int\frac{\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}$ $\text{I}_{1}=\int\frac{\text{2x+2}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}\text{dx}$ $x^2 + 2x + 3 = z^2$
(2x + 2)dx = 2z dz $\Rightarrow$ $\text{I}_{1}=\int\frac{\text{2z dz}}{\text{z}}$$=\text{2}\int{dz}=2z=2\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\text{C}_{1}$
$\Rightarrow$ $\text{I}_{1}=2\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\text{C}_{1}$ Again $\text{I}_{2}=\int\frac{\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}=\int\frac{\text{dx}}{\sqrt{\text{(x+1)}^{2}+(\sqrt{\text{2}}})^{2}}$$=\log|\text{(x+1)}+\sqrt{\text{(x+1)}^{2}+(\sqrt{\text{2}})^{2}}|$
$=\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{C}_{2}$
Putting the value of $I_1$, and $I_2$ in $(i)$ we get$\text{I}=\text{2}\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{(C}_{1}+\text{C}_{2})$
$=\text{2}\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{C.}$
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