Giải SBT Toán 12 Bài 11. Nguyên hàm

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

= 2x$^{3}$$\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) $\int {{{\left( {{e^x} - {e^{ - x}}} \right)}^2}dx}$ = $\int {\left( {{e^{2x}} - 2{e^x}.{e^{ - x}} + {e^{ - 2x}}} \right)} dx$

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

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

a) $\int {\left( {2\cos x + \frac{3}{{\sqrt x }}} \right)dx}$ = $\int {2\cos xdx + \int {\frac{3}{{\sqrt x }}} } dx$

= 2sinx + 6$\sqrt x$ + C.

b) $\int {\left( {3\sqrt x - 4\sin x} \right)dx}$ = $\int {3\sqrt x dx - \int {4\sin xdx} }$

= $2x\sqrt x + 4\cos x + C.$

a) $\int {\left( {x + {{\sin }^2}\frac{x}{2}} \right)dx}$ = $\int {xdx + \int {{{\sin }^2}\frac{x}{2}dx} }$

= $\int {xdx + \int {\frac{{1 - \cos x}}{2}dx} }$

= $\int {xdx + \int {\frac{1}{2}dx - \int {\frac{{\cos x}}{2}dx} } }$

= $\frac{1}{2}{x^2} + \frac{1}{2}x - \frac{1}{2}\sin x + C$.

b) $\int {{{\left( {2\tan x + \cot x} \right)}^2}dx}$ = $\int {\left( {4{{\tan }^2}x + 4\tan x\cot x + {{\cot }^2}x} \right)dx}$

= $\int {\left( {\frac{4}{{{{\cos }^2}x}} - 4 + 4 + \frac{1}{{{{\sin }^2}x}} - 1} \right)dx}$

= $\int {\frac{4}{{{{\cos }^2}xdx}} + \int {\frac{1}{{{{\sin }^2}x}}dx - \int {1dx} } }$

= 4tanx – cotx – x + C.

Độ cao h(t) của viên đạn tại điểm t là:

h(t) = $\int {\left( {150 - 9{,}8t} \right)dt}$ = 150t – 9,8$\frac{{{t^2}}}{2}$+ C = 150t – 4,9t$^{2}$ + C.

Thay t = 0 ta được h(0) = C = 0.

Vậy h(t) = 150t – 4,9t$^{2}$ (m).

a) Sau t = 3 giây, độ cao của viên đạn là:

h = h(3) = 150.3 – 4,9.3$^{2}$ = 405,9 (m).

b) Ta có: h(t) = 150t – 4,9t$^{2}$ (m).

h*'*(t) = v(t) = 150 – 9,8t

h*'*(t) = 0 ⇔ t = $\frac{{150}}{{9{,}8}}$.

Ta có bảng xét dấu như sau:

Khi đó, viên đạn đạt độ cao lớn nhất tại thời điểm t$_{max}$ = $\frac{{150}}{{9{,}8}}$.

Như vậy h$_{max}$ = 150t$_{max}$ – 4,9$t_{\max }^2$≈ 1148,0 (m).

Ta có: F*'*(u) = f(u), với mọi u ∈ K.

${\left[ {F\left( {u\left( x \right)} \right)} \right]^\prime }$ = $F'\left( {u\left( x \right)} \right)$.u*'*(x) = $f\left( {u\left( x \right)} \right).u'\left( x \right)$, với mọi x ∈ J.

Do đó, $\int {f\left( {u(x)} \right).u'(x)dx}$ = $F\left( {u\left( x \right)} \right)$ + C.

Áp dụng:

$\int {{{\left( {2x + 1} \right)}^5}dx}$ = ${\int {\left( {2x + 1} \right)} ^5}\frac{{{{\left( {2x + 1} \right)}^\prime }}}{2}dx$

= $\frac{1}{2}\int {{{\left( {2x + 1} \right)}^5}{{\left( {2x + 1} \right)}^\prime }dx}$

= $\frac{1}{2}.\frac{{{{\left( {2x + 1} \right)}^6}}}{6} + C$

= $\frac{{{{\left( {2x + 1} \right)}^6}}}{{12}} + C$.

$\int {\frac{1}{{\sqrt {2x + 1} }}dx}$ = $\int {\frac{1}{{\sqrt {2x + 1} }}.\frac{{{{\left( {2x + 1} \right)}^\prime }}}{2}dx}$

= $\int {\frac{1}{{2\sqrt {2x + 1} }}.{{\left( {2x + 1} \right)}^\prime }dx}$

= $\sqrt {2x + 1} + C.$

a) $\int {\frac{{2x - 1}}{{x + 1}}dx}$ = $\int {\left[ {\frac{{2\left( {x + 1} \right) - 3}}{{x + 1}}} \right]dx}$

= $\int {2dx - \int {\frac{3}{{x + 1}}dx} }$

= 2x – 3ln$\left| {x + 1} \right|$ + C.

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

= $\int {\left( {3 + 1 - \cos 2x} \right)dx}$

= $\int {\left( {4 - \cos 2x} \right)dx}$

= 4x − $\int {\cos 2x.\frac{{{{\left( {2x} \right)}^\prime }}}{2}dx}$

= 4x − $\frac{{\sin 2x}}{2}$ + C.