Giải bài 24 trang 15 sách bài tập toán 12 - Cánh diều

Đề bài

Tìm:

a) \(\int {{e^{5x}}} dx\);

b) \(\int {\frac{1}{{{{2024}^x}}}} dx\);

c) \(\int {\left( {{2^x} + {x^2}} \right)} dx\);

d) \(\int {\left( {{2^x}{{.3}^{2{\rm{x}} + 1}}} \right)} dx\);

e) \(\int {\frac{{{3^x} + {4^x} + 1}}{{{5^x}}}} dx\).

Lời giải chi tiết

a)

\(\int {{e^{5x}}} dx = \int {{{\left( {{e^5}} \right)}^x}} dx = \frac{{{{\left( {{e^5}} \right)}^x}}}{{\ln {e^5}}} + C = \frac{{{e^{5x}}}}{5} + C\).

b)

\(\int {\frac{1}{{{{2024}^x}}}} dx = \int {{{\left( {\frac{1}{{2024}}} \right)}^x}dx}  = \frac{{{{\left( {\frac{1}{{2024}}} \right)}^x}}}{{\ln \frac{1}{{2024}}}} + C =  - \frac{1}{{{{2024}^x}\ln 2024}} + C\).

c) \(\int {\left( {{2^x} + {x^2}} \right)dx}  = \frac{{{2^x}}}{{\ln 2}} + \frac{{{x^3}}}{3} + C\).

d) \(\int {\left( {{2^x}{{.3}^{2{\rm{x}} + 1}}} \right)dx}  = \int {\left( {{2^x}.{{\left( {{3^2}} \right)}^x}.3} \right)dx}  = \int {\left( {{2^x}{{.9}^x}.3} \right)dx}  = 3\int {{{\left( {2.9} \right)}^x}dx}  = 3\int {{{18}^x}dx}  = 3.\frac{{{{18}^x}}}{{\ln 18}} + C\).

e)

\(\begin{array}{l}\int {\frac{{{3^x} + {4^x} + 1}}{{{5^x}}}dx}  = \int {\left( {\frac{{{3^x}}}{{{5^x}}} + \frac{{{4^x}}}{{{5^x}}} + \frac{1}{{{5^x}}}} \right)dx}  = \int {\left( {{{\left( {\frac{3}{5}} \right)}^x} + {{\left( {\frac{4}{5}} \right)}^x} + {{\left( {\frac{1}{5}} \right)}^x}} \right)dx} \\ = \frac{{{{\left( {\frac{3}{5}} \right)}^x}}}{{\ln \frac{3}{5}}} + \frac{{{{\left( {\frac{4}{5}} \right)}^x}}}{{\ln \frac{4}{5}}} + \frac{{{{\left( {\frac{1}{5}} \right)}^x}}}{{\ln \frac{1}{5}}} + C = \frac{{{3^x}}}{{{5^x}\left( {\ln 3 - \ln 5} \right)}} + \frac{{{4^x}}}{{{5^x}\left( {\ln 4 - \ln 5} \right)}} - \frac{1}{{{5^x}\ln 5}} + C\end{array}\)