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Câu hỏi: Biết $I=\int\limits_{1}^{2}{\dfrac{dx}{\left( x+1 \right)\sqrt{x}+x\sqrt{x+1}}=\sqrt{a}-\sqrt{b}-c}$ với $a,b,c$ là các số nguyên dương. Tính $P=a+b+c.$
A. $P=24.$
B. $P=12.$
C. $P=18.$
D. $P=46.$
A. $P=24.$
B. $P=12.$
C. $P=18.$
D. $P=46.$
Ta có: $\sqrt{x+1}-\sqrt{x}\ne 0,\forall x\in \left[ 1;2 \right]$ nên:
$I=\int\limits_{1}^{2}{\dfrac{dx}{\left( x+1 \right)\sqrt{x}+x\sqrt{x+1}}}=\int\limits_{1}^{2}{\dfrac{dx}{\sqrt{x\left( x+1 \right)}\left( \sqrt{x+1}+\sqrt{x} \right)}}$
$=\int\limits_{1}^{2}{\dfrac{\left( \sqrt{x+1}-\sqrt{x} \right)dx}{\sqrt{x\left( x+1 \right)}\left( \sqrt{x+1}+\sqrt{x} \right)\left( \sqrt{x+1}-\sqrt{x} \right)}=\int\limits_{1}^{2}{\dfrac{\left( \sqrt{x+1}-\sqrt{x} \right)dx}{\sqrt{x}\left( x+1 \right)}}}$
$=\int\limits_{1}^{2}{\left( \dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}} \right)dx}=\left( 2\sqrt{x}-2\sqrt{x+1} \right)\left| _{\begin{smallmatrix}
\\
1
\end{smallmatrix}}^{\begin{smallmatrix}
2 \\
\end{smallmatrix}} \right.=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-2.$
Mà $I=\sqrt{a}-\sqrt{b}-c$ nên $\left\{ \begin{aligned}
& a=32 \\
& b=12 \\
& c=2 \\
\end{aligned} \right. $. Suy ra $ P=a+b+c=32+12+2=46.$
$I=\int\limits_{1}^{2}{\dfrac{dx}{\left( x+1 \right)\sqrt{x}+x\sqrt{x+1}}}=\int\limits_{1}^{2}{\dfrac{dx}{\sqrt{x\left( x+1 \right)}\left( \sqrt{x+1}+\sqrt{x} \right)}}$
$=\int\limits_{1}^{2}{\dfrac{\left( \sqrt{x+1}-\sqrt{x} \right)dx}{\sqrt{x\left( x+1 \right)}\left( \sqrt{x+1}+\sqrt{x} \right)\left( \sqrt{x+1}-\sqrt{x} \right)}=\int\limits_{1}^{2}{\dfrac{\left( \sqrt{x+1}-\sqrt{x} \right)dx}{\sqrt{x}\left( x+1 \right)}}}$
$=\int\limits_{1}^{2}{\left( \dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}} \right)dx}=\left( 2\sqrt{x}-2\sqrt{x+1} \right)\left| _{\begin{smallmatrix}
\\
1
\end{smallmatrix}}^{\begin{smallmatrix}
2 \\
\end{smallmatrix}} \right.=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-2.$
Mà $I=\sqrt{a}-\sqrt{b}-c$ nên $\left\{ \begin{aligned}
& a=32 \\
& b=12 \\
& c=2 \\
\end{aligned} \right. $. Suy ra $ P=a+b+c=32+12+2=46.$
Đáp án D.