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Let $x = \tan \theta$. Then $dx = \sec^2 \theta d\theta$. When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$. The integral becomes: $$I = \int_{0}^{\pi/4} \frac{\tan \theta \cdot \theta}{(\sec^2 \theta)^{3/2}} \cdot \sec^2 \theta d\theta = \int_{0}^{\pi/4} \frac{\theta \tan \theta}{\sec^3 \theta} \cdot \sec^2 \theta d\theta = \int_{0}^{\pi/4} \theta \sin \theta d\theta$$
Using the formula $\int u dv = uv - \int v du$, let $u = \theta$ and $dv = \sin \theta d\theta$. Then $du = d\theta$ and $v = -\cos \theta$. $$I = [\theta(-\cos \theta)]_{0}^{\pi/4} - \int_{0}^{\pi/4} (-\cos \theta) d\theta$$
$$I = [-\theta \cos \theta]_{0}^{\pi/4} + [\sin \theta]_{0}^{\pi/4}$$ $$I = (-\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} - 0) + (\frac{1}{\sqrt{2}} - 0) = \frac{1}{\sqrt{2}} - \frac{\pi}{4\sqrt{2}} = \frac{2-\pi}{4\sqrt{2}} = \frac{\sqrt{2}}{4} - \frac{\pi}{4\sqrt{2}}$$ Simplifying, we get $\frac{2-\pi}{4\sqrt{2}} = \frac{\sqrt{2}-\frac{\pi}{\sqrt{2}}}{4} = \frac{2-\pi}{4\sqrt{2}}$.
Final Answer: \frac{2-\pi}{4\sqrt{2}}
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