tính nguyên hàm bằng pp đổi biến số 11) I = `\int`(x². $e^{-x^3/3}$) dx 12) I = `\int`($\frac{e^x}{\sqrt[]{e^x+1}}$ )dx 13) I= `\int`($\frac{1}{e^x+2+e^{-x}}$ )dx 14) I = `\int`($\frac{1}{e^x+1}$ )dx

Đáp án:

$11)e^{-\tfrac{x^3}{3}}+C\\ 12)2\sqrt{e^x+1}+C\\ 13)-\dfrac{1}{e^x+1}+C\\ 14)x-\ln(e^x+1)+C.$

Giải thích các bước giải:

$11)\\ I=\displaystyle\int (x^2.e^{-\tfrac{x^3}{3}}) \, dx\\ t=-\dfrac{x^3}{3} \Rightarrow dt=-x^2 \, dx\\ I=-\displaystyle\int e^{t} \, dt\\ =-e^t+C\\ =e^{-\tfrac{x^3}{3}}+C\\ 12)\\ I=\displaystyle\int \dfrac{e^x}{\sqrt{e^x+1}} \, dx\\ t=e^x+1 \Rightarrow dt=e^x \, dx\\ I=\displaystyle\int \dfrac{dt}{\sqrt{t}} \\ =2\displaystyle\int \dfrac{dt}{2\sqrt{t}} \\ =2\sqrt{t}+C\\ =2\sqrt{e^x+1}+C\\ 13)\\ I=\displaystyle\int \dfrac{1}{e^x+2+e^{-x}} \, dx\\ =\displaystyle\int \dfrac{e^x}{(e^x)^2+2e^x+1} \, dx\\ =\displaystyle\int \dfrac{e^x}{(e^x+1)^2} \, dx\\ t=e^x+1 \Rightarrow dt=e^x \, dx\\ I=\displaystyle\int \dfrac{dt}{t^2}\\ =-\dfrac{1}{t}+C\\ =-\dfrac{1}{e^x+1}+C\\ 14)\\ I=\displaystyle\int \dfrac{1}{e^x+1} \, dx\\ =\displaystyle\int \dfrac{e^x}{e^x(e^x+1)} \, dx\\ t=e^x\Rightarrow dt=e^x \, dx\\ I=\displaystyle\int \dfrac{1}{t(t+1)} \, dt\\ =\displaystyle\int \left(\dfrac{1}{t}-\dfrac{1}{t+1}\right) \, dt\\ =\ln |t|-\ln|t+1|+C\\=\ln |e^x|-\ln|e^x+1|+C\\=x-\ln(e^x+1)+C.$