622225

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

© Được viết bởi CaolacVC. Blog https://caolacvc.blogspot.com

${\int\limits_0^1 \frac{4^x-1}{2^x-1} {~d} x=\int\limits_0^1 \frac{\left(2^x-1\right)\left(2^x+1\right)}{2^x-1} {~d} x=\int\limits_0^1\left(2^x+1\right) {d} x=\int\limits_0^1 2^x {~d} x+\int\limits_0^1 {~d} x}$

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