_0^4 1 x^2+2 x+3 d x - Vidyadip

Question

$\int_0^4 \frac{1}{\sqrt{x^2+2 x+3}} d x$

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Answer

$
\int_0^4 \frac{1}{\sqrt{x^2+2 x+3}} d x
$
$
\int_0^4 \frac{1}{\sqrt{\left(x^2+2 x+1\right)+2}} d x
$
$
\begin{aligned}
& \int_0^4 \frac{1}{\sqrt{(x+1)^2+(\sqrt{2})^2}} \\
& =\left[\log \left|(x+1)+\sqrt{\left.(x+1)^2+\sqrt{2}\right)^2}\right|\right]_0^4 \\
& =\left[\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|\right]_0^4 \\
& =\log (5+\sqrt{27})-\log (1+\sqrt{3}) \\
& =\log \left(\frac{5+3 \sqrt{3}}{1+\sqrt{3}}\right)
\end{aligned}
$

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