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If $f(x)=\sin \left(\tan ^{-1} x\right)$, then $\int_0^1 x f^{\prime \prime}(x) d x=$
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The correct answer is:
$-\frac{1}{2 \sqrt{2}}$
$\begin{aligned} f(x) & =\sin \left(\tan ^{-1} x\right) \\ f^{\prime}(x) & =\frac{\cos \left(\tan ^{-1} x\right)}{x^2+1} \\ I & =\int_0^1 x \cdot f^{\prime \prime}(x) d x={ }_0^1\left[x \cdot f^{\prime}(x)\right]-\int_0^1 f^{\prime}(x) d x \\ & =f^{\prime}(1)-{ }_0^1[f(x)]=\frac{1}{2 \sqrt{2}}-\frac{1}{\sqrt{2}} \\ & =-\frac{1}{2 \sqrt{2}}\end{aligned}$
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