$\int_{1}^{3}\frac{3+lnx}{(x+1)^2}dx$ Rãnh giúp giùm em.

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nguyenbahiep1

[laTEX]I = \int_{1}^{3}\frac{3dx}{(x+1)^2} + \int_{1}^{3}\frac{lnxdx}{(x+1)^2} =I_1+I_2\\ \\ I_1 = \frac{-3}{x+1} \big|_1^3 \\ \\ I_2 : u = ln x \Rightarrow du = \frac{1}{x} \\ \\ dv = \frac{1}{(x+1)^2} \Rightarrow v = \frac{-1}{(x+1)} \\ \\ I_2 = \frac{-lnx}{x+1} \big|_1^3 + \int_{1}^{3}\frac{dx}{x(x+1)} \\ \\ I_2 = \frac{-lnx}{x+1} \big|_1^3 + \int_{1}^{3}(\frac{1}{x} - \frac{1}{x+1})dx[/laTEX]
 
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