Giúp em bài tích phân

J

jet_nguyen

thangho12 said:
I=∫3223x2x2−1dxI=\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{2}}{3}} \dfrac{x^2}{\sqrt{x^2-1}}dx
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Gợi ý:
Mình tính nguyên hàm thôi nhé.
∫x2dxx2−1=∫(x2+1−1)dxx2−1=∫x2−1dx+∫dxx2−1\int {\dfrac{{{x^2}dx}}{{\sqrt {{x^2}- 1} }}} = \int {\dfrac{{\left( {{x^2} + 1 - 1} \right)dx}}{{\sqrt {{x^2} - 1} }}} = \int {\sqrt {{x^2} - 1} dx + } \int {\dfrac{{dx}}{{\sqrt {{x^2} - 1} }}}
=x2x2−1+12ln⁡∣x+x2−1∣+ln⁡∣x+x2−1∣+C = \dfrac{x}{2}\sqrt {{x^2} - 1} + \dfrac{1}{2}\ln \left| {x + \sqrt {{x^2} - 1} } \right| + \ln \left| {x + \sqrt {{x^2} - 1} } \right| + C
=12(xx2−1+3ln⁡∣x+x2−1∣)+C = \dfrac{1}{2}\left( {x\sqrt {{x^2} - 1} +3 \ln \left| {x + \sqrt {{x^2} - 1} } \right|} \right) + C

 
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