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Câu hỏi: Cho $\int\limits_{0}^{2}{\dfrac{x}{{{x}^{2}}+2\text{x}+4}d\text{x}}=a\ln 3+b\pi $ với a, b là các số thực. Giá trị của ${{a}^{2}}+3{{b}^{2}}$ bằng
A. $\dfrac{7}{27}$
B. $\dfrac{1}{2}$
C. $\dfrac{5}{18}$
D. $\dfrac{35}{144}$
A. $\dfrac{7}{27}$
B. $\dfrac{1}{2}$
C. $\dfrac{5}{18}$
D. $\dfrac{35}{144}$
Ta có: $\int\limits_{0}^{2}{\dfrac{x}{{{x}^{2}}+2\text{x}+4}d\text{x}}=\int\limits_{0}^{2}{\left( \dfrac{x+1}{{{x}^{2}}+2\text{x}+4}-\dfrac{1}{{{x}^{2}}+2\text{x}+4} \right)d\text{x}}$
$=\int\limits_{0}^{2}{\dfrac{x+1}{{{x}^{2}}+2\text{x}+4}d\text{x}}-\int\limits_{0}^{2}{\dfrac{1}{{{x}^{2}}+2\text{x}+4}d\text{x}}$.
Tính ${{I}_{1}}=\int\limits_{0}^{2}{\dfrac{x+1}{{{x}^{2}}+2\text{x}+4}d\text{x}}=\left. \dfrac{1}{2}\ln \left( {{x}^{2}}+2\text{x}+4 \right) \right|_{0}^{2}=\dfrac{1}{2}\left( \ln 12-\ln 4 \right)=\dfrac{1}{2}\ln 3$.
Tính ${{I}_{2}}=\int\limits_{0}^{2}{\dfrac{1}{{{x}^{2}}+2\text{x}+4}d\text{x}}=\int\limits_{0}^{2}{\dfrac{1}{{{\left( x+1 \right)}^{2}}+3}d\text{x}}$.
Đặt $x+1=\sqrt{3}\tan u\Rightarrow d\text{x}=\dfrac{\sqrt{3}}{{{\cos }^{2}}u}du$. Đổi cận: $x=0\Rightarrow u=\dfrac{\pi }{6}$ và $x=2\Rightarrow u=\dfrac{\pi }{3}$.
Suy ra ${{I}_{2}}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{\sqrt{3}}{{{\cos }^{2}}u}.\dfrac{1}{3\left( 1+{{\tan }^{2}}u \right)}du}=\dfrac{1}{\sqrt{3}}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{du}=\dfrac{1}{\sqrt{3}}\left( \dfrac{\pi }{3}-\dfrac{\pi }{6} \right)=\dfrac{\pi }{6\sqrt{3}}$.
Vậy $\int\limits_{0}^{2}{\dfrac{x}{{{x}^{2}}+2\text{x}+4}d\text{x}}={{I}_{1}}-{{I}_{2}}=\dfrac{1}{2}\ln 3-\dfrac{\pi }{6\sqrt{3}}$.
Suy ra ${{a}^{2}}+3{{b}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}+3.{{\left( \dfrac{1}{6\sqrt{3}} \right)}^{2}}=\dfrac{5}{18}$.
$=\int\limits_{0}^{2}{\dfrac{x+1}{{{x}^{2}}+2\text{x}+4}d\text{x}}-\int\limits_{0}^{2}{\dfrac{1}{{{x}^{2}}+2\text{x}+4}d\text{x}}$.
Tính ${{I}_{1}}=\int\limits_{0}^{2}{\dfrac{x+1}{{{x}^{2}}+2\text{x}+4}d\text{x}}=\left. \dfrac{1}{2}\ln \left( {{x}^{2}}+2\text{x}+4 \right) \right|_{0}^{2}=\dfrac{1}{2}\left( \ln 12-\ln 4 \right)=\dfrac{1}{2}\ln 3$.
Tính ${{I}_{2}}=\int\limits_{0}^{2}{\dfrac{1}{{{x}^{2}}+2\text{x}+4}d\text{x}}=\int\limits_{0}^{2}{\dfrac{1}{{{\left( x+1 \right)}^{2}}+3}d\text{x}}$.
Đặt $x+1=\sqrt{3}\tan u\Rightarrow d\text{x}=\dfrac{\sqrt{3}}{{{\cos }^{2}}u}du$. Đổi cận: $x=0\Rightarrow u=\dfrac{\pi }{6}$ và $x=2\Rightarrow u=\dfrac{\pi }{3}$.
Suy ra ${{I}_{2}}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{\sqrt{3}}{{{\cos }^{2}}u}.\dfrac{1}{3\left( 1+{{\tan }^{2}}u \right)}du}=\dfrac{1}{\sqrt{3}}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{du}=\dfrac{1}{\sqrt{3}}\left( \dfrac{\pi }{3}-\dfrac{\pi }{6} \right)=\dfrac{\pi }{6\sqrt{3}}$.
Vậy $\int\limits_{0}^{2}{\dfrac{x}{{{x}^{2}}+2\text{x}+4}d\text{x}}={{I}_{1}}-{{I}_{2}}=\dfrac{1}{2}\ln 3-\dfrac{\pi }{6\sqrt{3}}$.
Suy ra ${{a}^{2}}+3{{b}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}+3.{{\left( \dfrac{1}{6\sqrt{3}} \right)}^{2}}=\dfrac{5}{18}$.
Đáp án C.