Question
Find $\int \sqrt{x^{2}+2 x+5} d x$

Answer

Note that
$\int \sqrt{x^{2}+2 x+5} d x=\int \sqrt{(x+1)^{2}+4} d x$ 
Put x + 1 = y, so that dx = dy. Then
$\int \sqrt{x^{2}+2 x+5} d x=\int \sqrt{y^{2}+2^{2}} d y$ 
= $\frac{1}{2} y \sqrt{y^{2}+4}+\frac{4}{2} \log |y+\sqrt{y^{2}+4}|+C$ 
= $\frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5}+2 \log |x+1+\sqrt{x^{2}+2 x+5}|+\mathrm{C}$ 

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