622225

[622225] Tính$\underset{0}{\overset{1}{\int }}\frac{4^x-1}{2^x-1}dx$

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$\underset{0}{\overset{1}{\int }}\frac{4^x-1}{2^x-1}dx=\underset{0}{\overset{1}{\int }}\frac{(2^x-1)(2^x+1)}{2^x-1}dx=\underset{0}{\overset{1}{\int }}(2^x+1)dx=\underset{0}{\overset{1}{\int }}{2}^{x}dx+\underset{0}{\overset{1}{\int }}dx$

$={\frac{2^x}{\ln 2}|}_0^1+{x|}_0^1=\frac{2}{\ln 2}-\frac{1}{\ln 2}+1=1+\frac{1}{\ln 2}$