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Câu hỏi: Nếu $\int{\dfrac{{{e}^{x}}-1}{{{e}^{x}}+1}dx=2f(x)-x+C}$ thì $f(x)$ bằng
A. ${{e}^{x}}+1$.
B. ${{e}^{x}}$.
C. ${{e}^{x}}-1$.
D. $\ln \left( {{e}^{x}}+1 \right)$.
A. ${{e}^{x}}+1$.
B. ${{e}^{x}}$.
C. ${{e}^{x}}-1$.
D. $\ln \left( {{e}^{x}}+1 \right)$.
Ta có: $\int{\dfrac{{{e}^{x}}-1}{{{e}^{x}}+1}dx=\int{\left( 1-\dfrac{2}{{{e}^{x}}+1} \right)dx=\int{1dx-2\int{\dfrac{1}{{{e}^{x}}+1}dx}}}}$
Đặt: ${{e}^{x}}+1=u\Rightarrow {{e}^{x}}dx=du\Leftrightarrow dx=\dfrac{1}{u-1}du$
Nên: $\int{\dfrac{1}{{{e}^{x}}+1}dx=\int{\dfrac{1}{u\left( u-1 \right)}du=\int{\left( \dfrac{1}{u-1}-\dfrac{1}{u} \right)du=\ln \left| \dfrac{u-1}{u} \right|+{{C}_{1}}=\ln \dfrac{{{e}^{x}}}{{{e}^{x}}+1}+{{C}_{1}}}=x-\ln \left( {{e}^{x}}+1 \right)+{{C}_{1}}}}$
Vậy: $\int{\dfrac{{{e}^{x}}-1}{{{e}^{x}}+1}dx=x-2\left( x-\ln \left( {{e}^{x}}+1 \right)+{{C}_{1}} \right)=2\ln \left( {{e}^{x}}+1 \right)-x+C}$.
Đặt: ${{e}^{x}}+1=u\Rightarrow {{e}^{x}}dx=du\Leftrightarrow dx=\dfrac{1}{u-1}du$
Nên: $\int{\dfrac{1}{{{e}^{x}}+1}dx=\int{\dfrac{1}{u\left( u-1 \right)}du=\int{\left( \dfrac{1}{u-1}-\dfrac{1}{u} \right)du=\ln \left| \dfrac{u-1}{u} \right|+{{C}_{1}}=\ln \dfrac{{{e}^{x}}}{{{e}^{x}}+1}+{{C}_{1}}}=x-\ln \left( {{e}^{x}}+1 \right)+{{C}_{1}}}}$
Vậy: $\int{\dfrac{{{e}^{x}}-1}{{{e}^{x}}+1}dx=x-2\left( x-\ln \left( {{e}^{x}}+1 \right)+{{C}_{1}} \right)=2\ln \left( {{e}^{x}}+1 \right)-x+C}$.
Đáp án D.