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đặt u=3x+2ln(3x+1)$\Rightarrow du=3+\dfrac{6}{3x+1}$
dv=$\frac{1}{(x+1)^2}\Rightarrow v=\dfrac{-1}{x+1}$

=$\frac{-1}{x+1}$(3x+2ln(3x+1)$\frac{1}{0}$ -$\int_{0}^{1}\frac{-1}{x+1}((3+\frac{6}{3x+1})dx$

=$\frac{-3}{2}$-ln(4) +3$\int_{0}^{1}$($\frac{1}{x+1}$+$\frac{2}{(3x+1)(x+1)})$dx

=$\frac{-3}{2}$-ln(4)+3$\int_{0}^{1}$($\frac{1}{x+1}$)dx+3$\int_{0}^{1}$($\frac{3}{3x+1}$-$\frac{1}{x+1}$)dx

=$\frac{-3}{2}$-ln(4)+9$\int_{0}^{1}$($\frac{1}{3x+1})dx$

=$\frac{-3}{2}$-ln(4)+3ln(4)=$\frac{-3}{2}$+2ln(4)
 
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