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$\int_2^5 \sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} d x=$
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Verified Answer
The correct answer is:
$28 / 3$
$I=\int_q^5(\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}) d x$
$\begin{aligned} & \int_{\mathrm{q}}^5\left(\sqrt{(\mathrm{x}-1)^2+2 \sqrt{\mathrm{x}-1}+1}+\sqrt{(\mathrm{x}-1)^2-2 \sqrt{\mathrm{x}-1+1}}\right) \mathrm{dx} \\ & \int_{\mathrm{q}}^5[(\sqrt{\mathrm{x}-1}+1)+(\sqrt{\mathrm{x}-1}-1)] \mathrm{dx}\end{aligned}$
$\begin{aligned} & \int_{\mathrm{q}}^5(\mathrm{x}-1)^{1 / 2} \mathrm{dx}=\mathrm{q} \times \frac{2}{3}\left[(\mathrm{x}-1)^{3 / 2}\right]_2^5 \\ & \frac{4}{3} \times[8-1]=\frac{28}{3}\end{aligned}$
$\begin{aligned} & \int_{\mathrm{q}}^5\left(\sqrt{(\mathrm{x}-1)^2+2 \sqrt{\mathrm{x}-1}+1}+\sqrt{(\mathrm{x}-1)^2-2 \sqrt{\mathrm{x}-1+1}}\right) \mathrm{dx} \\ & \int_{\mathrm{q}}^5[(\sqrt{\mathrm{x}-1}+1)+(\sqrt{\mathrm{x}-1}-1)] \mathrm{dx}\end{aligned}$
$\begin{aligned} & \int_{\mathrm{q}}^5(\mathrm{x}-1)^{1 / 2} \mathrm{dx}=\mathrm{q} \times \frac{2}{3}\left[(\mathrm{x}-1)^{3 / 2}\right]_2^5 \\ & \frac{4}{3} \times[8-1]=\frac{28}{3}\end{aligned}$
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