Question

Let $S\left(x\right)=\int \frac{dx}{e^x+8e^{-x}+4e^{-3x}},R\left(x\right)=\int \frac{dx}{e^{3x}+8e^x+4e^{-x}}$ and $M\left(x\right)=S\left(x\right)-2R\left(x\right).$ If $M\left(x\right)=\frac{1}{2}ta{n}^{-1}\left(f\right(x\left)\right)+c$ then f(0)=

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

The correct option is A $\frac{3}{2}$

$M\left(x\right)=\int \frac{e^x(e^{2x}-2)}{e^{4x}+8e^{2x}+4}dxput{e}^x=t\Rightarrow \int \frac{(t^2-2)dt}{t^4+8t^2+4}=\frac{1}{2}ta{n}^{-1}\left(\frac{t+\frac{2}{t}}{2}\right)+c=\frac{1}{2}ta{n}^{-1}\left(\frac{e^x+2e^{-x}}{2}\right)+c$