Let f(x)=\int {e^{3 x}}/{4+8 e^{2 x}+e^{4 x}} d x and {g}(x)=\int {2 d x}/{e^{3 x}+8 e^{x}+4 e^{-x}} d x, then {f}(x)- {g}( {x})=

Correct Answer is : 12tan−1(ex+2e−x2)+C

\frac{1}{2} \tan ^{-1}( {e^{x}+2 e^{-x}}/{2})+C

\frac{1}{2} \tan ^{-1}( {e^{x}+e^{-x}}/{2})+C

\frac{1}{2} \tan ^{-1}( {2 e^{-x}+e^{-2 x}}/{2})+C

\frac{1}{2} \tan ^{-1}( {e^{2 x}+2 e^{x}}/{2 e^{-x}})+C

AP EAPCET 05-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

Let f(x)=∫‌

e3x

4+8e2x+e4x

dx and g(x)=∫‌

2dx

e3x+8ex+4e−x

dx, then f(x)−g(x)=

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