∫ f ( x ) d x = x^2/ 2 + 5 x − 7 ln x + C .

14/22

Cho hàm số \[f\left( x \right) = \frac{{{x^2} + 5x - 7}}{x}\].

a) \[\int {f\left( x \right)} \,{\rm{d}}x = \frac{{{x^2}}}{2} + 5x - 7\ln x + C\].

b) Hàm số \[f\left( x \right)\] là một nguyên hàm của hàm số \[g\left( x \right) = \frac{{{x^2} + 7}}{{{x^2}}}\].

c) \[\int\limits_{ - 2}^{ - 1} {f\left( x \right){\rm{d}}} x = \frac{m}{n} + m\ln n\], với \[m,n \in {\mathbb{N}^ * }\], \[\frac{m}{n}\] là phân số tối giản. Tổng \[m + 2025n = 4057\].

d) Gọi \[G\left( x \right)\] là một nguyên hàm của hàm số \[f\left( x \right)\] thỏa mãn \[G\left( 1 \right) = 4\] và \[G\left( 3 \right) + G\left( { - 9} \right) = 20\]. Khi đó \[G\left( { - 6} \right) = a\ln 2 + b\ln 3 + c\], với \[a,b,c\] là các số hữu tỉ. Tổng \[a + b + c = \frac{2}{3}\].

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

Ta có \[\int {f\left( x \right)} \,{\rm{d}}x = \int {\frac{{{x^2} + 5x - 7}}{x}} \,{\rm{d}}x = \int {\left( {x + 5 - \frac{7}{x}} \right)} \,{\rm{d}}x = \frac{{{x^2}}}{2} + 5x - 7\ln \left| x \right| + C\].

Ta thấy \[f'\left( x \right) = 1 + \frac{7}{{{x^2}}} = \frac{{{x^2} + 7}}{{{x^2}}} = g\left( x \right)\]. Do đó, \[f\left( x \right)\] là một nguyên hàm của hàm số \[g\left( x \right)\].

Ta có \[\int\limits_{ - 2}^{ - 1} {f\left( x \right){\rm{d}}} x = \left. {\left( {\frac{{{x^2}}}{2} + 5x - 7\ln \left| x \right|} \right)} \right|_{ - 2}^{ - 1} = \left( {\frac{1}{2} - 5} \right) - \left( {2 - 10 - 7\ln 2} \right) = \frac{7}{2} + 7\ln 2\].

Khi đó \[m = 7,\,\,n = 2\] nên \[m + 2025n = 4057\].

Ta có hàm số \[G\left( x \right)\] là một nguyên hàm của hàm số \[f\left( x \right)\] thì

\[G\left( x \right) = \frac{{{x^2}}}{2} + 5x - 7\ln \left| x \right| + C = \left\{ \begin{array}{l}\frac{{{x^2}}}{2} + 5x - 7\ln x + {C_1},\forall x \in \left( {0; + \infty } \right)\\\frac{{{x^2}}}{2} + 5x - 7\ln \left( { - x} \right) + {C_2},\forall x \in \left( { - \infty ;0} \right).\end{array} \right.\]

Ta có \[G\left( 1 \right) = 4 \Leftrightarrow \frac{1}{2} + 5 + {C_1} = 4 \Leftrightarrow {C_1} = - \frac{3}{2}\].

\[G\left( 3 \right) + G\left( { - 9} \right) = 20 \Leftrightarrow \frac{{{3^2}}}{2} + 5 \cdot 3 - 7\ln 3 - \frac{3}{2} + \frac{{{{\left( { - 9} \right)}^2}}}{2} + 5 \cdot \left( { - 9} \right) - 7\ln 9 + {C_2} = 20\]

\[ \Leftrightarrow {C_2} = \frac{{13}}{2} + 21\ln 3\].

Khi đó \[G\left( { - 6} \right) = \frac{{{{\left( { - 6} \right)}^2}}}{2} + 5.\left( { - 6} \right) - 7\ln 6 + \frac{{13}}{2} + 21\ln 3 = - 7\ln 2 + 14\ln 3 - \frac{{11}}{2}\].

Suy ra \[a = - 7,\,\,b = 14,\,\,c = - \frac{{11}}{2}\] nên \[a + b + c = \frac{3}{2}\].

Đáp án: a) Sai, b) Đúng, c) Đúng, d) Sai.