tích phân này làm sao ạ !

M

mrtom94

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S

sonsac99

minh nguyen ham ban the can

[TEX]I=\int\frac{ln(x^2+1)}{x^2}dx[/TEX]

[TEX]u=ln(x^2+1)\Rightarrow du=\frac{2x}{x^2+1}dx[/TEX]
[TEX]dv=\frac{1}{x^2}dx \Rightarrow v=\frac{-1}{x}[/TEX]
[TEX]I=\frac{-ln(x^2+1)}{x}+\int\frac{2}{x^2+1}dx[/TEX]
[TEX]x=tanu\Rightarrow dx=\frac{1}{cos^2u}du[/TEX]
[TEX]I=\frac{-ln(x^2+1)}{x}+\int 2du[/tex]
[TEX]I=\frac{-ln(x^2+1)}{x}+2u[/TEX]
 
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