Giải bài 2 trang 25 sách bài tập toán 12 - Chân trời sáng tạo

Đề bài

Tìm:

a) \(\int {{{\left( {3{\rm{x}} - \frac{1}{{{x^2}}}} \right)}^2}dx} \); b) \(\int {\left( {7{\rm{x}}\sqrt[3]{x} - \frac{1}{{\sqrt {{x^3}} }}} \right)dx} \left( {x > 0} \right)\);

c) \(\int {{{\left( {{3^{2{\rm{x}}}} - 1} \right)}^2}dx} \);    d) \(\int {\left( {2 - 3{{\cos }^2}\frac{x}{2}} \right)dx} \).

Lời giải chi tiết

a)

\(\begin{array}{l}\int {{{\left( {3{\rm{x}} - \frac{1}{{{x^2}}}} \right)}^2}dx}  = \int {\left( {9{{\rm{x}}^2} - \frac{6}{x} + \frac{1}{{{x^4}}}} \right)dx}  = \int {\left( {9{{\rm{x}}^2} - 6.\frac{1}{x} + {x^{ - 4}}} \right)dx} \\ = 3{{\rm{x}}^3} - 6\ln \left| x \right| + \frac{{{x^{ - 3}}}}{{ - 3}} + C = 3{{\rm{x}}^3} - 6\ln \left| x \right| - \frac{1}{{3{{\rm{x}}^3}}} + C\end{array}\)

b) \(\int {\left( {7{\rm{x}}\sqrt[3]{x} - \frac{1}{{\sqrt {{x^3}} }}} \right)dx}  = \int {\left( {7{{\rm{x}}^{\frac{4}{3}}} - {x^{ - \frac{3}{2}}}} \right)dx}  = 7\frac{{{{\rm{x}}^{\frac{7}{3}}}}}{{\frac{7}{3}}} - \frac{{{x^{ - \frac{1}{2}}}}}{{ - \frac{1}{2}}} + C = 3{{\rm{x}}^2}\sqrt[3]{x} + \frac{2}{{\sqrt x }} + C\).

c)

\(\begin{array}{l}\int {{{\left( {{3^{2{\rm{x}}}} - 1} \right)}^2}dx}  = \int {\left( {{3^{4{\rm{x}}}} - {{2.3}^{2{\rm{x}}}} + 1} \right)dx}  = \int {\left( {{{81}^{\rm{x}}} - {{2.9}^{\rm{x}}} + 1} \right)dx}  = \frac{{{{81}^{\rm{x}}}}}{{\ln 81}} - 2.\frac{{{9^{\rm{x}}}}}{{\ln 9}} + x + C\\ = \frac{{{3^{4{\rm{x}}}}}}{{4\ln 3}} - \frac{{{3^{{\rm{2x}}}}}}{{\ln 3}} + x + C\end{array}\)

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