Simple Definite Integral

I am presenting 2 approaches to this problem, it's upto the reader which one to follow.

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Approach 1

Let $\displaystyle \mathcal{I}=\int_{0}^{1}\frac{1-x}{\left(x+1\right)\sqrt{x^{3}+x^{2}+x}}~\mathrm{d}x$ . Now multilplying up and down by $x+1$ and dividing by $x^2$ results in $\mathcal{I}=-\int_{0}^{1}\frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}~\mathrm{d}x$

Now letting $u=x+\frac{1}{x}$ transforms the integral to $\mathcal{I}=\int_{2}^{\infty}\frac{\mathrm{d}u}{\left(u+2\right)\sqrt{u+1}}$

Now the substitution $u+1=t^2$ makes $\mathcal{I}=2\int_{\sqrt{3}}^{\infty}\frac{\mathrm{d}t}{1+t^{2}}$ which is simply $2(\tan^{-1}{\infty}-\tan^{-1}{\sqrt{3}})=2\left(\dfrac{\pi}{2}-\dfrac{\pi}{3}\right)=\boxed{\dfrac{\pi}{3}}$ .

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Approach 2

Let $x=\tan^2\theta$ , and therefore $\mathrm{d}x=2\tan\theta\sec^2\theta$ , we get

$\begin{aligned} \mathcal{I} &=2\int_{0}^{\frac{\pi}{4}}\frac{1-\tan^{2}\theta}{\sqrt{1+\tan^{2}\theta+\tan^{4}\theta}}~\mathrm{d}\theta \\ &=2\int_{0}^{\frac{\pi}{4}}\frac{\cos2\theta}{\sqrt{\sin^{4}\theta+\sin^{2}\theta\cos^{2}\theta+\cos^{4}\theta}}~\mathrm{d}\theta \\ &=2\int_{0}^{\frac{\pi}{4}}\frac{\cos2\theta}{\sqrt{\left(\sin^{2}\theta+\cos^{2}\theta\right)^{2}-\sin^{2}\theta\cos^{2}\theta}}~\mathrm{d}\theta \\ &=2\int_{0}^{\frac{\pi}{4}}\frac{\cos2\theta}{\sqrt{1-\left(\dfrac{\sin2\theta}{2}\right)^{2}}} \, \mathrm{d}\theta \\ &=2\int_{0}^{\frac{1}{2}}\frac{\mathrm{d}u}{\sqrt{1-u^2}} \, ~\text{via}~\frac{\sin2\theta}{2}=u \\ &=2\sin^{-1}u \bigr]_{0}^{\frac{1}{2}} \\ &=2\sin^{-1}\left(\frac{1}{2}\right) \\ &=\boxed{\dfrac{\pi}{3}} \end{aligned}$