Giải SBT Toán 12 Tập 2 KNTT Bài 11. Nguyên hàm có đáp án

Ta có: f(x) = \(\int {f'\left( x \right)dx} \)

                  = \(\int {\left( {3\sqrt x + \frac{2}{{\sqrt[3]{x}}}} \right)} dx\)

                  = \(\int {3\sqrt x dx + \int {\frac{2}{{\sqrt[3]{x}}}} } dx\)

                  = 2x\(\sqrt x \) + 3\(\sqrt[3]{{{x^2}}}\) + C.

Mà f(1) = 1 nên 2 + 3 + C = 1 hay C = −4.

Vậy f(x) = 2x\(\sqrt x \) + 3\(\sqrt[3]{{{x^2}}}\) − 4.

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

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

                          = \(\int {\frac{1}{{{x^2}}}dx + \int {\frac{2}{{{x^3}}}dx + \int {\frac{1}{{{x^4}}}dx} } } \)

                          = \( - \frac{1}{x}\) + 4.\(\frac{{{x^{ - 2}}}}{{ - 2}}\) + 4.\(\frac{{{x^{ - 3}}}}{{ - 3}}\) + C

                          = \( - \frac{1}{x}\) − \(\frac{2}{{{x^2}}}\) − \(\frac{4}{{3{x^3}}}\) + C.

b) \(\int {\sqrt x \left( {7{x^2} + 6} \right)dx} \) = \(\int {\left( {7{x^2}\sqrt x + 6\sqrt x } \right)} dx\)

                                 = \(\int {7{x^2}\sqrt x dx + \int {6\sqrt x } } dx\)

                                 = \(7\int {{x^{\frac{5}{2}}}} dx + 6\int {{x^{\frac{1}{2}}}dx} \)

                                 = 2\({x^{\frac{7}{2}}} + 4{x^{\frac{3}{2}}}\) + C

                                 = 2x³\(\sqrt x \) + 4x\(\sqrt x \) + C.

a) \(\int {\left( {3x + 4} \right)\sqrt[3]{x}} dx\) = \(\int {\left( {3x\sqrt[3]{x} + 4\sqrt[3]{x}} \right)} dx\)

                               = \(\int {3x\sqrt[3]{x}dx + \int {4\sqrt[3]{x}dx} } \)

                               = \(\int {3{x^{\frac{4}{3}}}dx} \) + \(\int {4{x^{\frac{1}{3}}}} dx\)

                               = \(\frac{9}{7}{x^2}\sqrt[3]{x} + 3x\sqrt[3]{x}\) + C

                               = \(\left( {\frac{9}{7}{x^2} + 3x} \right)\sqrt[3]{x}\) + C

b) \(\int {\frac{{{{\left( {2x + 3} \right)}^2}}}{{\sqrt x }}dx} \) = \(\int {\frac{{4{x^2} + 12x + 9}}{{\sqrt x }}dx} \)

                            = \(\int {\left( {4x\sqrt x + 12\sqrt x + \frac{9}{{\sqrt x }}} \right)dx} \)

                            = \(\int {4x\sqrt x dx + \int {12\sqrt x } } dx + \int {\frac{9}{{\sqrt x }}dx} \)

                            = \(4\int {{x^{\frac{3}{2}}}} dx + 12\int {{x^{\frac{1}{2}}}dx + 9\int {{x^{\frac{{ - 1}}{2}}}dx} } \)

                            = \(\frac{8}{5}{x^2}\sqrt x + 8x\sqrt x + 18\sqrt x + C\)

                            = \(\left( {\frac{8}{5}{x^2} + 8x + 18} \right)\sqrt x \) + C.

a) \(\int {\left( {2{e^x} + \frac{1}{{{3^x}}}} \right)} dx\) = \(\int {2{e^x}dx + \int {\frac{1}{{{3^x}}}dx} } \)

                              = \(\int {2{e^x}dx + \int {{3^{ - x}}dx} } \)

                              = 2ex − \(\frac{1}{{{3^x}.\ln 3}}\) + C.

b) \(\int {\left( {{x^2} + {2^x}} \right)} dx\) = \(\int {{x^2}dx + \int {{2^x}dx} } \)

                           = \(\frac{{{x^3}}}{3} + \frac{{{2^x}}}{{\ln 2}} + C\).

a) \(\int {{{\left( {{2^x} + {3^x}} \right)}^2}dx} \) = \(\int {\left( {{2^{2x}} + {{2.6}^x} + {3^{2x}}} \right)dx} \)

                            = \(\int {\left( {{4^x} + {{2.6}^x} + {9^x}} \right)} dx\)

                            = \(\int {{4^x}dx + \int {{{2.6}^x}dx + \int {{9^x}dx} } } \)

                            = \(\frac{{{4^x}}}{{\ln 4}} + \frac{{{{2.6}^x}}}{{\ln 6}} + \frac{{{9^x}}}{{\ln 9}}\)+ C.

b) \(\int {{{\left( {{e^x} - {e^{ - x}}} \right)}^2}dx} \) = \(\int {\left( {{e^{2x}} - 2{e^x}.{e^{ - x}} + {e^{ - 2x}}} \right)} dx\)

                              = \(\int {{e^{2x}}dx + \int {{e^{ - 2x}}dx - \int {2dx} } } \)

                              = \(\frac{{{e^{2x}}}}{2} - \frac{{{e^{ - 2x}}}}{2} - 2x + C\).