Giải bài 59 trang 29 sách bài tập toán 12 - Cánh diều

Đề bài

a) \(\int {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)dx} \);

b) \(\int {x\left( {2 - \frac{3}{{{x^3}}}} \right)dx} \);

c) \(\int {{e^{ - 3{\rm{x}}}}dx} \);

d) \(\int {\left( {2 - 3{{\tan }^2}x} \right)dx} \);

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

g) \(\int {\frac{{{3^{2{\rm{x}} + 1}}}}{{{2^x}}}dx} \).

Lời giải chi tiết

a) \(\int {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)dx}  = \int {\left( {{x^3} + 1} \right)dx}  = \frac{{{x^4}}}{4} + x + C\).

b) \(\int {x\left( {2 - \frac{3}{{{x^3}}}} \right)dx}  = \int {\left( {2x - 3{x^{ - 2}}} \right)dx}  = {x^2} - 3.\frac{{{x^{ - 1}}}}{{ - 1}} + C = {x^2} + \frac{3}{x} + C\).

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

d) \(\int {\left( {2 - 3{{\tan }^2}x} \right)dx}  = \int {\left[ {2 - 3\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)} \right]dx}  = \int {\left[ {5 - 3.\frac{1}{{{{\cos }^2}x}}} \right]dx}  = 5{\rm{x}} - 3\tan x + C\).

e) \(\int {\frac{1}{{{2^{ - x + 1}}}}dx}  = \int {\frac{1}{{{2^{ - x}}.2}}dx}  = \int {\frac{1}{2}.{2^x}dx}  = \frac{1}{2}.\frac{{{2^x}}}{{\ln 2}} + C = \frac{{{2^x}}}{{2\ln 2}} + C\).

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