Google
Câu hỏi: Biết $\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+1}+\left( x+1 \right)\sqrt{x}}=\sqrt{a}-\sqrt{b}-\sqrt{c}}$ với $a,b,c$ là các số nguyên dương. Tính $P=a+b+c.$
A. $P=44.$
B. $P=42.$
C. $P=46.$
D. $P=48.$
A. $P=44.$
B. $P=42.$
C. $P=46.$
D. $P=48.$
Đặt $I=\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+1}+\left( x+1 \right)\sqrt{x}}=\int\limits_{1}^{2}{\dfrac{dx}{\sqrt{x\left( x+1 \right)}\left( \sqrt{x}+\sqrt{x+1} \right)}.}}$
Đặt $t=\sqrt{x}+\sqrt{x+1}\Rightarrow dt=\dfrac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x\left( x+1 \right)}}dx\Leftrightarrow \dfrac{dx}{\sqrt{x\left( x+1 \right)}}=2\dfrac{dt}{t}.$
Khi $x=1$ thì $t=\sqrt{2}+1$, khi $x=2$ thì $t=\sqrt{3}+\sqrt{2}.$
$I=\int\limits_{1}^{2}{\dfrac{dx}{\sqrt{x\left( x+1 \right)}\left( \sqrt{x}+\sqrt{x+1} \right)}=2\int\limits_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}}{\dfrac{dt}{{{t}^{2}}}}=-2\dfrac{1}{t}\left| _{\begin{smallmatrix}
\\
\sqrt{2}+1
\end{smallmatrix}}^{\begin{smallmatrix}
\sqrt{3}+\sqrt{2} \\
\end{smallmatrix}} \right.=-2\left( \dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{2}+1} \right)}$
$=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-\sqrt{4}\Rightarrow a=32;b=12;c=4\Rightarrow P=48.$
Đặt $t=\sqrt{x}+\sqrt{x+1}\Rightarrow dt=\dfrac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x\left( x+1 \right)}}dx\Leftrightarrow \dfrac{dx}{\sqrt{x\left( x+1 \right)}}=2\dfrac{dt}{t}.$
Khi $x=1$ thì $t=\sqrt{2}+1$, khi $x=2$ thì $t=\sqrt{3}+\sqrt{2}.$
$I=\int\limits_{1}^{2}{\dfrac{dx}{\sqrt{x\left( x+1 \right)}\left( \sqrt{x}+\sqrt{x+1} \right)}=2\int\limits_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}}{\dfrac{dt}{{{t}^{2}}}}=-2\dfrac{1}{t}\left| _{\begin{smallmatrix}
\\
\sqrt{2}+1
\end{smallmatrix}}^{\begin{smallmatrix}
\sqrt{3}+\sqrt{2} \\
\end{smallmatrix}} \right.=-2\left( \dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{2}+1} \right)}$
$=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-\sqrt{4}\Rightarrow a=32;b=12;c=4\Rightarrow P=48.$
Đáp án D.