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Câu hỏi: Biết rằng tích phân $\int_0^4 \dfrac{(x+1) e^x}{\sqrt{2 x+1}} d x=a e^4+b$. Tính $T=a^2-b^2$
A. $T=1$.
B. $T=2$.
C. $T=\dfrac{3}{2}$.
D. $T=\dfrac{5}{2}$.
A. $T=1$.
B. $T=2$.
C. $T=\dfrac{3}{2}$.
D. $T=\dfrac{5}{2}$.
Ta có $I=\int_0^4 \dfrac{x+1}{\sqrt{2 x+1}} e^x d x=\dfrac{1}{2} \int_0^4 \dfrac{2 x+2}{\sqrt{2 x+1}} e^x d x=\dfrac{1}{2}\left(\int_0^4 \sqrt{2 x+1} \cdot e^x d x+\int_0^4 \dfrac{e^x}{\sqrt{2 x+1}} d x\right)$.
Xét $I_1=\int_0^4 \dfrac{e^x}{\sqrt{2 x+1}} d x$.
Đặt $\left\{\begin{array}{l}u=e^x \\ d v=\dfrac{d x}{\sqrt{2 x+1}}\end{array} \Rightarrow\left\{\begin{array}{l}d u=e^x d x \\ v=\int \dfrac{d x}{\sqrt{2 x+1}}=\dfrac{1}{2} \cdot \dfrac{(2 x+1)^{\dfrac{1}{2}}}{\dfrac{1}{2}}=\sqrt{2 x+1}\end{array}\right.\right.$
Do đó $I_1=\left.e^x \cdot \sqrt{2 x+1}\right|_0 ^4-\int_0^4 e^x \cdot \sqrt{2 x+1} d x$.
Suy ra $I=\dfrac{3 e^4-1}{2}$. Khi đó $a=\dfrac{3}{2}, b=\dfrac{-1}{2} \Rightarrow T=\dfrac{9}{4}-\dfrac{1}{4}=2$.
Xét $I_1=\int_0^4 \dfrac{e^x}{\sqrt{2 x+1}} d x$.
Đặt $\left\{\begin{array}{l}u=e^x \\ d v=\dfrac{d x}{\sqrt{2 x+1}}\end{array} \Rightarrow\left\{\begin{array}{l}d u=e^x d x \\ v=\int \dfrac{d x}{\sqrt{2 x+1}}=\dfrac{1}{2} \cdot \dfrac{(2 x+1)^{\dfrac{1}{2}}}{\dfrac{1}{2}}=\sqrt{2 x+1}\end{array}\right.\right.$
Do đó $I_1=\left.e^x \cdot \sqrt{2 x+1}\right|_0 ^4-\int_0^4 e^x \cdot \sqrt{2 x+1} d x$.
Suy ra $I=\dfrac{3 e^4-1}{2}$. Khi đó $a=\dfrac{3}{2}, b=\dfrac{-1}{2} \Rightarrow T=\dfrac{9}{4}-\dfrac{1}{4}=2$.
Đáp án B.