88. $$I = \int^1_\frac12 \sqrt{\dfrac{x}{x^3+1}} \, \mathrm{d}x$$
Đặt $t = \sqrt{\dfrac{x^3}{x^3+1}}$ thì $\dfrac1{t^2} = 1 + \dfrac{1}{x^3}$
suy ra $-\dfrac{2}{t^3} \, \mathrm{d}t = -\dfrac{3}{x^4} \, \mathrm{d}x$
và $x^3 = \dfrac{t^2}{1 - t^2}$
$$\begin{align} I &= \int^1_\frac12 \dfrac{x^3}{3} \cdot \sqrt{\dfrac{x^3}{x^3 + 1}} \cdot \dfrac{3}{x^4} \, \mathrm{d}x \\
&= \int^\sqrt{\frac12}_\frac13 \dfrac{t^2}{3(1-t^2)} \cdot t \cdot \dfrac{2}{t^3} \, \mathrm{d}t \\
&= \int^\sqrt{\frac12}_\frac13 \dfrac{2}{3(1-t^2)} \, \mathrm{d}t \\
&= \int^\sqrt{\frac12}_\frac13 \left[ \dfrac{1}{3(1+t)} + \dfrac{1}{3(1-t)} \right] \, \mathrm{d}t \\
&= \left.\left( \dfrac13 \ln |1 + t| - \dfrac13 \ln |1 - t| \right)\right|^\sqrt\frac12_\frac13 \\
&= \left. \dfrac13 \ln\dfrac{|1 + t|}{|1 - t|}\right|^\sqrt\frac12_\frac13 \\
&= \dfrac13 \ln (3 + 2\sqrt{2}) - \dfrac13 \ln 2 \\
&= \dfrac13 \ln (\dfrac{3}2 + \sqrt{2})\end{align}$$
Vậy $a = 3$, $b = 3$, $c = 2$, $d = 2$ nên tổng $= 10$. Chọn B
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