nguyên hàm của e^3x(2e^x+e^-x+1)

Đáp án:

$\dfrac{1}{2}e^{4x}+\dfrac{1}{2} e^{2x}+ \dfrac{1}{3} e^{3x}+C.$

Giải thích các bước giải:

$\displaystyle\int e^{3x}(2e^x+e^{-x}+1) \, dx\\ =\displaystyle\int (2e^{4x}+e^{2x}+e^{3x}) \, dx\\ =\displaystyle\int 2e^{4x}\, dx+\displaystyle\int e^{2x} \, dx + \displaystyle\int e^{3x} \, dx\\ =\dfrac{1}{2}\displaystyle\int e^{4x}\, d(4)x+\dfrac{1}{2}\displaystyle\int e^{2x} \, d(2)x + \dfrac{1}{3}\displaystyle\int e^{3x} \, d(3x)\\ =\dfrac{1}{2}e^{4x}+\dfrac{1}{2} e^{2x}+ \dfrac{1}{3} e^{3x}+C.$