Tích phân \(\int\limits^2_1\frac{\ln\left(1+x\right)}{x^2}\text{d}x\) bằng
\(3\ln\frac{2\sqrt{3}}{3}\). \(2\ln\frac{\sqrt{3}}{3}\). \(3\ln2\). \(3\ln\sqrt{3}\). Hướng dẫn giải:Đặt \(\begin{cases}u=\ln\left(1+x\right)\\v'=\dfrac{1}{x^2}\end{cases}\) suy ra \(\begin{cases}u'=\dfrac{1}{1+x}\\v=\int x^{-2}\text{d}x=-x^{-1}=-\dfrac{1}{x}\end{cases}\)
\(I=-\dfrac{\ln\left(1+x\right)}{x}|^2_1+\int\limits^2_1\dfrac{1}{x\left(x+1\right)}\text{d}x\)
Ta có: \(\int\dfrac{1}{x\left(x+1\right)}\text{d}x=\int\left[\dfrac{1}{x}-\dfrac{1}{x+1}\right]\text{d}x=\ln x-\ln\left(x+1\right)\)
Suy ra:
\(I=\left[-\dfrac{\ln\left(1+x\right)}{x}+\ln x-\ln\left(1+x\right)\right]|^2_1\)
\(=-\dfrac{3}{2}\ln3+3\ln2=3\ln\dfrac{2}{\sqrt{3}}=3\ln\dfrac{2\sqrt{3}}{3}\)
Đáp số : \(I=3\ln\dfrac{2\sqrt{3}}{3}\).
Cách khác: Dùng MTCT kiểm tra các đáp số.