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Integrate the function
$\frac{x+2}{\sqrt{x^2+2 x+3}}$
$\frac{x+2}{\sqrt{x^2+2 x+3}}$
Solution:
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Verified Answer
$\mathrm{I}=\frac{1}{2} \int \frac{2 x+4}{\sqrt{x^2+2 x+3}} d x$
$=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
$=\mathrm{I}_1+\mathrm{I}_2+\mathrm{C}($ Let $)$
$\mathrm{I}_1=\frac{1}{2} \int \frac{d t}{\sqrt{t}}=\frac{1}{2} \times 2 t^{\frac{1}{2}}=\sqrt{x^2+2 x+3}$
$\mathrm{I}_2=\int \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}}=\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|$
$\therefore \mathrm{I}=\sqrt{x^2+2 x+3}+\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|+\mathrm{C}$
$=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
$=\mathrm{I}_1+\mathrm{I}_2+\mathrm{C}($ Let $)$
$\mathrm{I}_1=\frac{1}{2} \int \frac{d t}{\sqrt{t}}=\frac{1}{2} \times 2 t^{\frac{1}{2}}=\sqrt{x^2+2 x+3}$
$\mathrm{I}_2=\int \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}}=\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|$
$\therefore \mathrm{I}=\sqrt{x^2+2 x+3}+\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|+\mathrm{C}$
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