Khẳng định nào sau đây sai?
\(\int\limits\dfrac{x\text{dx}}{3-2x^2}=-\dfrac{1}{4}\ln\left|3-2x^2\right|+C\). \(\int\dfrac{\text{dx}}{x\ln x}=\int\dfrac{\text{d}\ln x}{\ln x}=\ln\left|\ln x\right|+C\).\(\int\limits\dfrac{\text{dx}}{1+\cos x}=\dfrac{1}{2}\tan\dfrac{x}{2}+C\).\(\int\limits\dfrac{\text{dx}}{x\sqrt{x^2+1}}=\dfrac{1}{2}\ln\left|\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}+1}\right|+C\).Hướng dẫn giải:Ta có \(\int\dfrac{x\text{dx}}{3-2x^2}=\int\dfrac{-\frac{1}{4}\text{d}\left(2x^2-3\right)}{2x^2-3}=-\dfrac{1}{4}\ln\left|2x^2-3\right|+C\).
Tương tự \(\int\dfrac{\text{d}x}{x\ln x}=\int\dfrac{\text{d}\ln x}{\ln x}=\ln\left|\ln x\right|+C\)
\(\int\limits\dfrac{\text{dx}}{1+\cos x}=\int\dfrac{\text{dx}}{2\cos^2\dfrac{x}{2}}=\int\dfrac{\text{d}\left(\frac{x}{2}\right)}{\cos^2\dfrac{x}{2}}=\tan\dfrac{x}{2}+C\), suy ra \(\int\limits\dfrac{\text{dx}}{1+\cos x}=\dfrac{1}{2}\tan\dfrac{x}{2}+C\) là khẳng định sai.
