Bài 113005

Question

Tính tích phân : $J= \int\limits_{0}^{1}x\ln (x^2+x+1)dx $

score 0 · Answer 1

Đặt : $\left\{ \begin{array}{l} u=\ln (x^2+x+1) \Rightarrow du=\frac{2x+1}{x^2+x+1}dx \\ dv=xdx \Rightarrow v=\frac{x^2}{2} \end{array} \right. $
$J= \frac{x^2}{2}\ln (x^2+x+1)|^1_0-\int\limits_{0}^{1}\frac{x^3+\frac{1}{2}x^2 }{x^2+x+1} $
$=\frac{1}{2}\ln 3-\int\limits_{0}^{1} \left ( x-\frac{1}{2} +\frac{1}{2} .\frac{-x+1}{x^2+x+1} \right ) dx $
$=\frac{1}{2} \ln 3-[\frac{x^2}{2} -\frac{1}{2} x]|^1_0+\frac{1}{4}\int\limits_{0}^{1}\frac{2x+1-3}{x^2+x+1} dx $
$=\frac{1}{2}\ln 3-0+\frac{1}{4}\int\limits_{0}^{1}\frac{d(x^2+x+1)}{x^2+x+1}-\frac{3}{4}\int\limits_{0}^{1}\frac{dx}{(x+\frac{1}{2} )^2+(\frac{{\sqrt{3}} }{2} )^2} $
$=\frac{1}{2} \ln 3+\frac{1}{4} \ln (x^2+x+1)|^1_0-\frac{3}{4} .\frac{1}{\frac{{\sqrt{3}} }{2} } \arctan \frac{x+\frac{1}{2} }{\frac{{\sqrt{3}} }{2} } |^1_0$
$\Rightarrow J=\frac{1}{2} \ln 3+\frac{1}{4} \ln 3-\frac{{\sqrt{3}} }{2} \left ( \arctan {\sqrt{3}}-\arctan \frac{1}{{\sqrt{3}} } \right ) $
$=\frac{3}{4}\ln 3-\frac{{\sqrt{3}} }{2} (\frac{\pi}{3} -\frac{\pi}{6}) =\frac{3}{4} \ln 3-\frac{{\sqrt{3}}\pi }{12} (ycbt)$

Bài 113005

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Bài 113005

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