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maxmin · 04-09-2011, 10:28 PM

Bài 24: Tính các tích phân sau:
$\[\begin{array}{l}
a)\int\limits_0^1 {\frac{{{x^{2001}}dx}}{{{{\left( {{x^2} + 1} \right)}^{1002}}}}} \\
b)\int\limits_0^1 {\frac{{{x^3}dx}}{{{{\left( {{x^8} - 4} \right)}^2}}}} \\
c)\int\limits_0^1 {\frac{{\left( {2{x^2} + 5x - 2} \right)dx}}{{{x^3} + 2{x^2} - 4x - 8}}} \\
d)\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\sqrt {t{g^2}x + \cot {g^2}x - 2} dx} \\
e)\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{dx}}{{\sin x.\sin \left( {x + \frac{\pi }{6}} \right)}}} \\
f)\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x + 7\cos x + 6}}{{4\sin x + 3\cos x + 5}}dx} \\
\end{array}\] $
( Nguồn diendantoanhoc.net )

04-09-2011, 10:28 PM	#38
maxmin +Thành Viên+ Tham gia ngày: Mar 2009 Bài gởi: 304 Thanks: 70 Thanked 142 Times in 89 Posts	Bài 24: Tính các tích phân sau: $\[\begin{array}{l} a)\int\limits_0^1 {\frac{{{x^{2001}}dx}}{{{{\left( {{x^2} + 1} \right)}^{1002}}}}} \\ b)\int\limits_0^1 {\frac{{{x^3}dx}}{{{{\left( {{x^8} - 4} \right)}^2}}}} \\ c)\int\limits_0^1 {\frac{{\left( {2{x^2} + 5x - 2} \right)dx}}{{{x^3} + 2{x^2} - 4x - 8}}} \\ d)\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\sqrt {t{g^2}x + \cot {g^2}x - 2} dx} \\ e)\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{dx}}{{\sin x.\sin \left( {x + \frac{\pi }{6}} \right)}}} \\ f)\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x + 7\cos x + 6}}{{4\sin x + 3\cos x + 5}}dx} \\ \end{array}\] $ ( Nguồn diendantoanhoc.net )