Tính: a) 2 ∫ 1 x 4 + x 3 + x 2 + x + 1 x 2 d x ; b) 2 ∫ 1 x e x + 1 x d x ; c) 1 ∫ 0 8 x + 1 2 x + 1 d x ; d) π 2 ∫ π 4 1 + sin 2 x 1 − cos 2 x d x .
a) Ta có: \[\int\limits_1^2 {\frac{{{x^4} + {x^3} + {x^2} + x + 1}}{{{x^2}}}dx} \]
\[ = \int\limits_1^2 {\left( {{x^2} + x + 1 + \frac{1}{x} + \frac{1}{{{x^2}}}} \right)dx} \]
\[ = \left. {\left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + \ln \left| x \right| - \frac{1}{x}} \right)} \right|_1^2\]
\[ = \ln 2 + \frac{{16}}{3}.\]
b) Ta có: \[\int\limits_1^2 {\frac{{x{e^x} + 1}}{x}dx} = \int\limits_1^2 {\left( {{e^x} + \frac{1}{x}} \right)dx} \]
\[ = \left. {\left( {{e^x} + \ln \left| x \right|} \right)} \right|_1^2\]
= e2 − e + ln2.
c) Ta có:
\[\int\limits_0^1 {\frac{{{8^x} + 1}}{{{2^x} + 1}}dx} = \int\limits_0^1 {\frac{{\left( {{2^x} + 1} \right)\left( {{4^x} - {2^x} + 1} \right)}}{{\left( {{2^x} + 1} \right)}}dx} \]
\[ = \int\limits_0^1 {\left( {{4^x} - {2^x} + 1} \right)dx} \]
\[ = \left. {\left( {\frac{{{4^x}}}{{\ln 4}} - \frac{{{2^x}}}{{\ln 2}} + x} \right)} \right|_0^1\]
\[ = \frac{4}{{\ln 4}} - \frac{2}{{\ln 2}} + 1 - \frac{1}{{\ln 4}} + \frac{1}{{\ln 2}}\]
\[ = 1 + \frac{1}{{2\ln 2}}\].
d) Ta có:
\[\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{1 + {{\sin }^2}x}}{{1 - {{\cos }^2}x}}dx} = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{1 + {{\sin }^2}x}}{{{{\sin }^2}x}}dx} \]
\[ = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {\frac{1}{{{{\sin }^2}x}} + 1} \right)dx} \]
\[ = \left. {\left( { - \cot x + x} \right)} \right|_{_{\frac{\pi }{4}}}^{\frac{\pi }{2}} = 1 + \frac{\pi }{4}\].