[tex]I=\int\limits_{1}^{3}\frac{ln(x^2 +3)}{x^2}dx[/tex]
bài giải
đặt
[TEX]\left{\begin{u=ln(x^2+3)}\\{dv=\frac{1}{x^2}dx} [/TEX]
[TEX]\left{\begin{du=\frac{2x}{x^2+3}dx}\\{v=\frac{-1}{x}} [/TEX]
[TEX]I = \frac{- ln(x^2 +3)}{x}\mid_{1}^{3} +\int_{1}^{3} \frac{2}{x^2+3}dx[/TEX]
theo yêu cầu bạn thuỷ tôi giải tiếp
[TEX]\int_{1}^{3} \frac{2}{x^2+3}dx[/TEX]
đặt[TEX] x = \sqrt{3} tant =>dx= \sqrt{3} ( 1+tan^2t) dt[/TEX]
[TEX]x\in [1;3] =>t \in [\frac{\pi}{6};\frac{\pi}{3} ][/TEX]
[TEX]\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{2\sqrt{3}}{3} dt [/TEX]