Giải bài 3 trang 25 sách bài tập toán 12 - Chân trời sáng tạo

Đề bài

Tính:

a) \(\int\limits_1^2 {\frac{{{x^4} + {x^3} + {x^2} + x + 1}}{{{x^2}}}dx} \);

b) \(\int\limits_1^2 {\frac{{x{e^x} + 1}}{x}dx} \);

c) \(\int\limits_0^1 {\frac{{{8^x} + 1}}{{{2^x} + 1}}dx} \);

d) \(\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{1 + {{\sin }^2}x}}{{1 - {{\cos }^2}x}}dx} \).

Lời giải chi tiết

a) \(\int\limits_1^2 {\frac{{{x^4} + {x^3} + {x^2} + x + 1}}{{{x^2}}}dx}  = \int\limits_1^2 {\left( {{x^2} + x + 1 + \frac{1}{x} + {x^{ - 2}}} \right)dx}  = \left. {\left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + \ln \left| x \right| - \frac{1}{x}} \right)} \right|_1^2 = \ln 2 + \frac{{16}}{3}\).

b) \(\int\limits_1^2 {\frac{{x{e^x} + 1}}{x}dx}  = \int\limits_1^2 {\left( {{e^x} + \frac{1}{x}} \right)dx}  = \left. {\left( {{e^x} + \ln \left| x \right|} \right)} \right|_1^2 = {e^2} - e + \ln 2\).

c)

\(\begin{array}{l}\int\limits_0^1 {\frac{{{8^x} + 1}}{{{2^x} + 1}}dx}  = \int\limits_0^1 {\frac{{{2^{3x}} + 1}}{{{2^x} + 1}}dx}  = \int\limits_0^1 {\frac{{\left( {{2^x} + 1} \right)\left( {{2^{2x}} - {2^x} + 1} \right)}}{{{2^x} + 1}}dx}  = \int\limits_0^1 {\left( {{4^x} - {2^x} + 1} \right)dx} \\ = \left. {\left( {\frac{{{4^x}}}{{\ln 4}} - \frac{{{2^x}}}{{\ln 2}} + x} \right)} \right|_0^1 = 1 + \frac{1}{{2\ln 2}}\end{array}\)

d) \(\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{1 + {{\sin }^2}x}}{{1 - {{\cos }^2}x}}dx}  = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{1 + {{\sin }^2}x}}{{{{\sin }^2}x}}dx}  = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {\frac{1}{{{{\sin }^2}x}} + 1} \right)dx}  = \left. {\left( { - \cot x + x} \right)} \right|_{\frac{\pi }{4}}^{\frac{\pi }{2}} = 1 + \frac{\pi }{4}\).