Chủ đề này có 2 trả lời

[bleuceiu]

tìm nguyên hàm $\int x^{2}\sqrt{x^{2}+1}dx$

[huutuant10]

https://diendantoanh...phân-chebyshev/

[chanhquocnghiem]

tìm nguyên hàm $\int x^{2}\sqrt{x^{2}+1}dx$

$I=\int  x^2\sqrt{x^2+1}\ dx$

Đặt $x=\frac{e^t-e^{-t}}{2}=\frac{e^{2t}-1}{2e^t}\Rightarrow e^{2t}-2xe^t-1=0\Rightarrow e^t=\sqrt{x^2+1}+x\Rightarrow t=\ln(\sqrt{x^2+1}+x)$

$\sqrt{x^2+1}=\sqrt{\frac{e^{2t}+2+e^{-2t}}{4}}=\frac{e^t+e^{-t}}{2}$

$dx=\frac{e^t+e^{-t}}{2}\ dt$

$I=\int \left [ \frac{(e^t-e^{-t})^2}{4}.\frac{(e^t+e^{-t})^2}{4} \right ]\ dt=\frac{1}{16}\int (e^{2t}-e^{-2t})^2dt$

$=\frac{1}{16}\int (e^{4t}+e^{-4t}-2)dt=\frac{e^{4t}-e^{-4t}-8t}{64}+C$

$=\frac{e^{4\ln(\sqrt{x^2+1}+x)}-e^{-4\ln(\sqrt{x^2+1}+x)}-8\ln(\sqrt{x^2+1}+x)}{64}+C$.