Câu hỏi:
72. Họ các nguyên hàm của hàm số \(f\left( x \right) = \frac{{{{\left( {x – 2021} \right)}^{2020}}}}{{{{\left( {x + 2022} \right)}^{2022}}}}\) là
A. \(\frac{1}{{2022}}{\left( {\frac{{x – 2021}}{{x + 2022}}} \right)^{2020}} + C\). B.\(\frac{{2021}}{{4043}}{\left( {\frac{{x – 2021}}{{x + 2022}}} \right)^{2021}} + C\).
C. \(\frac{1}{{4043.2021}}{\left( {\frac{{x – 2021}}{{x + 2022}}} \right)^{2021}} + C\). D.\(\frac{1}{{2021}}.\frac{{4043}}{{{{\left( {x + 2022} \right)}^{2021}}}} + C\).
Lời giải
Ta có \(I = \int {f\left( x \right)} {\rm{d}}x = \int {\frac{{{{\left( {x – 2021} \right)}^{2020}}}}{{{{\left( {x + 2022} \right)}^{2022}}}}{\rm{d}}x} = \int {{{\left( {\frac{{x – 2021}}{{x + 2022}}} \right)}^{2020}}\frac{1}{{{{\left( {x + 2022} \right)}^2}}}{\rm{d}}x} \)
Đặt \(t = \frac{{x – 2021}}{{x + 2022}} \Rightarrow {\rm{d}}t = \frac{{4043}}{{{{\left( {x + 2022} \right)}^2}}}{\rm{d}}x \Rightarrow \frac{1}{{4043}}{\rm{d}}t = \frac{1}{{{{\left( {x + 2022} \right)}^2}}}{\rm{d}}x\).
Suy ra: \(I = \frac{1}{{4043}}\int {{t^{2020}}{\rm{d}}t = \frac{1}{{4043.2021}}{t^{2021}} + C} \).
\( \Rightarrow \)\(\int {f\left( x \right)} {\rm{d}}x = \frac{1}{{4043.2021}}{\left( {\frac{{x – 2021}}{{x + 2022}}} \right)^{2021}} + C\).
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Thuộc chủ đề: Trắc nghiệm Tích phân
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