Biết \(\int\limits_{0}^{4} \frac{\sqrt{2 x+1} \mathrm{d} x}{2 x+3 \sqrt{2 x+1}+3}=a+b \ln 2+c \ln \frac{5}{3}(a, b, c \in \mathbb{Z}) . \text { Tính } T=2 a+b+c\)
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Lời giải:
Báo saiTa có:
\(I=\int\limits_{0}^{4} \frac{\sqrt{2 x+1} \mathrm{d} x}{2 x+3 \sqrt{2 x+1}+3}=\int\limits_{0}^{4} \frac{\sqrt{2 x+1} \mathrm{d} x}{(\sqrt{2 x+1}+1)(\sqrt{2 x+1}+2)}=\int\limits_{0}^{4} \frac{2(\sqrt{2 x+1}+1)-(\sqrt{2 x+1}+2) \mathrm{d} x}{(\sqrt{2 x+1}+1)(\sqrt{2 x+1}+2)}\)
\(=\int\limits_{0}^{4} \frac{2 \mathrm{d} x}{(\sqrt{2 x+1}+2)}-\int\limits_{0}^{4} \frac{\mathrm{d} x}{(\sqrt{2 x+1}+1)}\)
Đặt \(u=\sqrt{2 x+1} \Rightarrow u \mathrm{d} u=\mathrm{d} x . \text { Với } x=0 \Rightarrow u=1, \text { với } x=4 \Rightarrow u=3\)
\(I=\int\limits_{1}^{3} \frac{2 u \mathrm{d} u}{u+2}-\int\limits_{1}^{3} \frac{u \mathrm{d} u}{u+1}=\int\limits_{1}^{3}\left(2-\frac{4}{u+2}\right) \mathrm{d} u-\int\limits_{1}^{3}\left(1-\frac{1}{u+1}\right) \mathrm{d} u\)
\(\begin{array}{l} =\left.(u-4 \ln |u+2|+\ln |u+1|)\right|_{1} ^{3}=2-4 \ln \frac{5}{3}+\ln 2 \\ \Rightarrow a=2, b=1, c=1 \Rightarrow T=2.1+1-4=1 \end{array}\)
