\(\int\limits^1_0e^{3x+1}\text{d}x\) bằng
\(\dfrac{1}{3}\left(e^4-e\right)\).\(e^4-e\).\(e^3-e\).\(\dfrac{1}{3}\left(e^4+e\right)\).Hướng dẫn giải:\(\int e^{3x+1}\text{d}x=\frac{1}{3}\int e^{3x+1}\text{d}\left(3x+1\right)=\frac{1}{3}e^{3x+1}+C\). Do đó \(\int\limits^1_0e^{3x+1}\text{d}x=\frac{1}{3}\left(e^4-e\right)\).