<s>[INST] $\int_{0}^{\frac{\sqrt{2}-1}{2}}\frac{dx}{(2x+1)\sqrt{x^2+x}}$ $\int_{0}^{\frac{\sqrt{2}-1}{2}}\frac{dx}{(2x+1)\sqrt{x^2+x}}$
This is in the form of $\frac{1}{linear\sqrt{quadratic}}$.I put $x=\frac{1}{t}$
$\int_{\frac{2}{\sqrt2-1}}^{\infty}\frac{dt}{(2+t)\sqrt{t+1}}$Then put $t+1=p^2$
From now,it got complicated.Its answer is $\frac{\pi}{4}$.Answer is elusive.
 [/INST] HINT:
As $x^2+x=\dfrac{(2x+1)^2-1^2}4$
start with $2x+1=\sec\theta$
See: Trigonometric substitutions
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