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Câu hỏi: Tìm nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)={{3}^{x}}\sqrt{{{3}^{x}}+1}$
A. $F\left( x \right)=\dfrac{{{3}^{x}}\left( 2+{{3}^{x+1}} \right)\ln 3}{2\sqrt{{{3}^{x}}+1}}$
B. $F\left( x \right)=\dfrac{2}{3}.\left( {{3}^{x}}+1 \right)\sqrt{{{3}^{x}}+1}+C$
C. $F\left( x \right)=\dfrac{2\sqrt{{{3}^{x}}+1}}{3\ln 3}+C$
D. $F\left( x \right)=\dfrac{2\left( {{3}^{x}}+1 \right)\sqrt{{{3}^{x}}+1}}{3\ln 3}+C$
A. $F\left( x \right)=\dfrac{{{3}^{x}}\left( 2+{{3}^{x+1}} \right)\ln 3}{2\sqrt{{{3}^{x}}+1}}$
B. $F\left( x \right)=\dfrac{2}{3}.\left( {{3}^{x}}+1 \right)\sqrt{{{3}^{x}}+1}+C$
C. $F\left( x \right)=\dfrac{2\sqrt{{{3}^{x}}+1}}{3\ln 3}+C$
D. $F\left( x \right)=\dfrac{2\left( {{3}^{x}}+1 \right)\sqrt{{{3}^{x}}+1}}{3\ln 3}+C$
Lời giải:
Ta có
$\int{{{3}^{x}}}\sqrt{{{3}^{x}}+1}dx=\dfrac{1}{\ln 3}\int{\sqrt{{{3}^{x}}+1}(}{{3}^{x}})=\dfrac{1}{\ln 3}\int{{{({{3}^{x}}+1)}^{\dfrac{1}{2}}}({{3}^{x}}+1)=\dfrac{1}{\ln 3}\dfrac{2\sqrt{{{({{3}^{x}}+1)}^{3}}}}{3}+C=\dfrac{2({{3}^{x}}+1)\sqrt{{{3}^{x}}+1}}{3\ln 3}+C.}$
Ta có
$\int{{{3}^{x}}}\sqrt{{{3}^{x}}+1}dx=\dfrac{1}{\ln 3}\int{\sqrt{{{3}^{x}}+1}(}{{3}^{x}})=\dfrac{1}{\ln 3}\int{{{({{3}^{x}}+1)}^{\dfrac{1}{2}}}({{3}^{x}}+1)=\dfrac{1}{\ln 3}\dfrac{2\sqrt{{{({{3}^{x}}+1)}^{3}}}}{3}+C=\dfrac{2({{3}^{x}}+1)\sqrt{{{3}^{x}}+1}}{3\ln 3}+C.}$
Đáp án D.