$\int_{0}^{\frac{\sqrt{2}-1}{2}}\frac{dx}{(2x+1)\sqrt{x^2+x}}$

Question

$\int_{0}^{\frac{\sqrt{2}-1}{2}}\frac{dx}{(2x+1)\sqrt{x^2+x}}$

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.

Answer 1

HINT:

As $x^2+x=\dfrac{(2x+1)^2-1^2}4$

start with $2x+1=\sec\theta$

See: Trigonometric substitutions

Answer 2

$$\int_{2/(\sqrt2-1)}^{\infty}\frac{dt}{(2+t)\sqrt{t+1}}$$

Let $t+1=u^2\implies dt=2udu$ $$\int_{\sqrt 2-1}^{\infty}\frac{2udu}{(1+u^2)u}$$ $$=2\int_{\sqrt 2-1}^{\infty}\frac{du}{1+u^2}$$ $$=2\left[\tan^{-1}(u)\right]_{\sqrt 2-1}^{\infty}$$ $$=2\left[\tan^{-1}(\infty)-\tan^{-1}\left(\sqrt 2-1\right)\right]$$ $$=2\left[\frac{\pi}{2}-\frac{\pi}{8}\right]$$ $$=2\frac{3\pi}{8}$$$$=\frac{3\pi}{4}$$

Answer 3

Let $\sqrt{x^2+x}=u$

$\implies\dfrac{2x+1}{2\sqrt{x^2+x}}dx=du$ and $(2x+1)^2=4u^2+1$

$$\implies\int\dfrac{dx}{(2x+1)\sqrt{x^2+x}}=\int\dfrac{(2x+1)dx}{(2x+1)^2\sqrt{x^2+x}}=\int\dfrac{2du}{4u^2+1}=?$$