Google
Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=\dfrac{2\sqrt{x+3}}{2\sqrt{x+3}+x}$ sau phép đặt $t=\sqrt{x+3}$ là
A. $F\left( t \right)=4t+\ln \left| t-1 \right|-9\ln \left| t+3 \right|+C$.
B. $F\left( t \right)=4t-\ln \left| t+1 \right|+9\ln \left| t-3 \right|+C$.
C. $F\left( t \right)=4t-\ln \left| t-1 \right|+9\ln \left| t+3 \right|+C$.
D. $F\left( t \right)=4t+\ln \left| t+1 \right|-9\ln \left| t-3 \right|+C$.
A. $F\left( t \right)=4t+\ln \left| t-1 \right|-9\ln \left| t+3 \right|+C$.
B. $F\left( t \right)=4t-\ln \left| t+1 \right|+9\ln \left| t-3 \right|+C$.
C. $F\left( t \right)=4t-\ln \left| t-1 \right|+9\ln \left| t+3 \right|+C$.
D. $F\left( t \right)=4t+\ln \left| t+1 \right|-9\ln \left| t-3 \right|+C$.
Đặt $t=\sqrt{x+3}\Leftrightarrow {{t}^{2}}=x+3\Rightarrow 2tdt=dx$ khi đó ta có
$\int{\dfrac{2\sqrt{x+3}}{2\sqrt{x+3}+x}dx=\int{\dfrac{2t.2tdt}{{{t}^{2}}+2t-3}=\int{\dfrac{4\left( {{t}^{2}}+2t-3 \right)+\left( t+3 \right)-9\left( t-1 \right)}{\left( t+3 \right)\left( t-1 \right)}dt}}}$
$=\int{\left( 4+\dfrac{1}{t-1}-\dfrac{9}{t+3} \right)dt=4t+\ln \left| t-1 \right|-9\ln \left| t+3 \right|+C}$
$\int{\dfrac{2\sqrt{x+3}}{2\sqrt{x+3}+x}dx=\int{\dfrac{2t.2tdt}{{{t}^{2}}+2t-3}=\int{\dfrac{4\left( {{t}^{2}}+2t-3 \right)+\left( t+3 \right)-9\left( t-1 \right)}{\left( t+3 \right)\left( t-1 \right)}dt}}}$
$=\int{\left( 4+\dfrac{1}{t-1}-\dfrac{9}{t+3} \right)dt=4t+\ln \left| t-1 \right|-9\ln \left| t+3 \right|+C}$
Đáp án A.