giải chi tiết giúp e với ạ

Đáp án: $\mathcal{A.}\; \dfrac{-\ln{2x} - x \ln{2}}{x+1}+ \ln{ \left| \dfrac{x}{x+1} \right|} + 1999$

Giải thích các bước giải:

Gọi $\mathcal{I}= \displaystyle \int \dfrac{\ln{x}}{(x+1)^2}\text{d}x$

Chọn $\displaystyle \left \{ {{u=\ln{x}} \atop {\text{d}v= \dfrac{1}{(x+1)^2}\text{d}x}} \right. \Leftrightarrow \left \{ {{\text{d} u=\dfrac{\text{d}x}{x}} \atop {v=\dfrac{-1}{x+1} }} \right.$ 

$\Longrightarrow \mathcal{I} = \ln{x} \cdot \dfrac{-1}{x+1} - \displaystyle \int \dfrac{-1}{x+1} \cdot \dfrac{1}{x} \text{d}x$

$\Longleftrightarrow \mathcal{I} = - \dfrac{\ln{x}}{x+1} + \displaystyle \int \dfrac{\text{d}x}{x(x+1)}$

$\Longleftrightarrow \mathcal{I} = - \dfrac{\ln{x}}{x+1} + \displaystyle \int \dfrac{\text{d}x}{x}  - \displaystyle \int \dfrac{\text{d}x}{x+1}$

$\Longleftrightarrow \mathcal{I} = - \dfrac{\ln{x}}{x+1} + \ln{|x|} - \ln{|x+1|}+ \mathcal{C} \left(\mathcal{C} \in \mathbb{R} \right)$

$\Longleftrightarrow \mathcal{I} = - \dfrac{\ln{x}}{x+1} + \ln{ \left| \dfrac{x}{x+1} \right|} + \mathcal{C} \;\;\;\;\; \left(\mathcal{C} \in \mathbb{R} \right)$

Chọn $\mathcal{C}=1999-\ln{2}$

$\Longrightarrow \mathcal{I} = - \dfrac{\ln{x}}{x+1} + \ln{ \left| \dfrac{x}{x+1} \right|} + 1999 - \ln{2}$

$\Longleftrightarrow \mathcal{I} = - \dfrac{\ln{x}}{x+1}-\dfrac{\ln{2}\cdot(x+1)}{x+1} + \ln{ \left| \dfrac{x}{x+1} \right|} + 1999$

$\Longleftrightarrow \mathcal{I} = \dfrac{-\ln{x} -x \ln{2}-\ln{2}}{x+1}+ \ln{ \left| \dfrac{x}{x+1} \right|} + 1999$

$\Longleftrightarrow \mathcal{I} = -\dfrac{\ln{2}+\ln{x}+x\ln{2}}{x+1}+ \ln{ \left| \dfrac{x}{x+1} \right|} + 1999$

$\Longleftrightarrow \mathcal{I} = -\dfrac{\ln{2x}+x\ln{2}}{x+1}+ \ln{ \left| \dfrac{x}{x+1} \right|} + 1999$

$\Longrightarrow$ Chọn đáp án. $\mathcal{A.}$

$@thomasnguyen364$