Giải SBT Toán 12 Chân trời sáng tạo Bài 1. Nguyên hàm có đáp án

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

$$= \int {{x^2}dx - \int {4xdx + \int {4dx} } }$$

$= \frac{{{x^3}}}{3} - 2{x^2} + 4x + C$.

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

$$= \int {3{x^2}dx - \int {2xdx - \int {1dx} } }$$

= x3 – x2 + x + C.

c) $\int {\sqrt[3]{{{x^2}}}dx} = \int {{x^{\frac{2}{3}}}dx = \frac{3}{5}{x^{\frac{5}{3}}} + C = \frac{3}{5}x\sqrt[3]{{{x^2}}}} + C.$

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

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

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

$$= 2{x^{\frac{1}{2}}} - 2.\frac{2}{3}{x^{\frac{3}{2}}} + \frac{2}{5}{x^{\frac{5}{2}}} + C$$

$$= 2\sqrt x - \frac{4}{3}x\sqrt x + \frac{2}{5}{x^2}\sqrt x + C.$$

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

$$= \int {{5^{2x}}dx - \int {1dx = \int {{{25}^x}dx} - \int {1dx} } }$$

$$= \frac{{{{25}^x}}}{{\ln 25}} - x + C = \frac{{{{25}^x}}}{{2\ln 5}} - x + C.$$

b) $\int {{e^{ - 0{,}5x}}dx} = \int {{{\left( {{e^{ - 0{,}5}}} \right)}^x}dx = \frac{{{{\left( {{e^{ - 0{,}5}}} \right)}^x}}}{{\ln {e^{ - 0{,}5}}}}} + C$

$= \frac{{{e^{ - 0{,}5x}}}}{{ - 0{,}5}} + C = - 2{e^{ - 0{,}5x}} + C$.

c) $\int {{2^{x - 1}}{{.5}^{2x + 1}}dx} = \int {\frac{{{2^x}}}{2}{{.5}^{2x}}.5dx = \int {\frac{5}{2}{{.2}^x}{{.25}^x}dx} }$

$$= \frac{5}{2}\int {{{50}^x}dx = \frac{5}{2}.\frac{{{{50}^x}}}{{\ln 50}} + C.}$$

a)$\int {\frac{{{{\cos }^2}x}}{{1 - \sin {\rm{x}}}}dx} = \int {\frac{{1 - {{\sin }^2}x}}{{1 - \sin {\rm{x}}}}dx}$

$$= \int {\frac{{\left( {1 - \sin {\rm{x}}} \right)\left( {1 + \sin {\rm{x}}} \right)}}{{\left( {1 - \sin {\rm{x}}} \right)}}dx}$$

$$= \int {\left( {1 + \sin {\rm{x}}} \right)dx} = x - \cos x + C.$$

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

$$= \int {\left( {\frac{5}{2} - \frac{3}{2}\cos x} \right)dx}$$

$= \frac{5}{2}x - \frac{3}{2}\sin {\rm{x}} + C$.

c) $\int {\frac{{2{{\cos }^3}x + 3}}{{{{\cos }^2}x}}dx} = \int {\left( {2\cos x + \frac{3}{{{{\cos }^2}x}}} \right)dx}$

$$= \int {2\cos xdx + \int {\frac{3}{{{{\cos }^2}x}}dx} }$$

$$= 2\sin x + 3\tan x + C$$

a) $f\left( x \right) = \int {f'\left( x \right)dx}$

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

= $\frac{{{x^4}}}{2} - 2{x^2} + x + C.$

Mà f(1) = 0 ⇒ $\frac{1}{2} - 2 + 1 + C = 0$⇒ $C = \frac{1}{2}$.

Vậy $f\left( x \right) = \frac{{{x^4}}}{2} - 2{x^2} + x + \frac{1}{2}.$

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

$$= \int {\left( {5\cos x - \sin {\rm{x}}} \right)dx}$$

$= 5\sin x + \cos x + C$.

Mà $f\left( {\frac{\pi }{2}} \right) = 1$ nên $5\sin \left( {\frac{\pi }{2}} \right) + \cos \left( {\frac{\pi }{2}} \right) + C = 1$ hay $5 + C = 1$ suy ra C = −4.

Vậy f(x) = 5sinx + cosx – 4.

Theo giả thiết, hệ số góc của tiếp tuyến tại mỗi điểm (x; f(x)) là $\frac{{1 - x}}{{{x^2}}}$ với x > 0 hay $f'\left( x \right) = \frac{{1 - x}}{{{x^2}}}$ với x > 0 và f(1) = 2.

Ta có: $f\left( x \right) = \int {f'\left( x \right)} dx = \int {\frac{{1 - x}}{{{x^2}}}dx}$

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

Mà f(1) = 2 nên −1 – ln1 + C = 2 hay C = 3.

Vậy $f\left( x \right) - \frac{1}{x} - \ln x + 3.$