Chủ đề này có 2 trả lời

[luuquangtrung68]

1)$\int \frac{dx}{(10e^{-x}-1)(\sqrt{e^{x}-1})}$

2) $\int \frac{x.dx}{\sqrt{x^{2}+1+\sqrt{(x^{2}+1)^{3}}}}$
3) $\int \frac{(log_{2}x)^{3}dx}{x.\sqrt{1+3ln^{2}x}}$

 

[lanhmacluongbac]

1)$\int \frac{dx}{(10e^{-x}-1)(\sqrt{e^{x}-1})}$

 

Đặt $t=\sqrt{e^{x}-1}\Rightarrow dt= \frac{e^x}{2\sqrt{e^x-1}}dx$

Thay vào bt ta có:

$\int \frac{dx}{(10e^{-x}-1)(\sqrt{e^{x}-1})} =\int \frac{2dt}{9-t^2} = \frac{1}{3}\int (\frac{1}{t-3}-\frac{1}{t+3})dt =\frac{1}{3}\ln \left | \frac{t-3}{t+3} \right |+c =\frac{1}{3}\ln \left | \frac{\sqrt{e^{x}-1}-3}{\sqrt{e^{x}-1}+3} \right |+c$

[lanhmacluongbac]

2) $\int \frac{x.dx}{\sqrt{x^{2}+1+\sqrt{(x^{2}+1)^{3}}}}$

 

 $I=\int \frac{x.dx}{\sqrt{x^{2}+1+\sqrt{(x^{2}+1)^{3}}}}=\int \frac{xdx}{\sqrt{x^2+1}.\sqrt{1+\sqrt{x^2+1}}}$

Đặt $t=\sqrt{x^2+1}\Rightarrow dt=\frac{x}{\sqrt{x^2+1}}dx\Leftrightarrow tdt=xdx$

$\Rightarrow I=\int \frac{dt}{\sqrt{1+t}}=2\sqrt{1+t}+c= 2\sqrt{1+\sqrt{x^2+1}}+c$