Question

Let $I_1=\int \frac{dx}{e^x+8e^{-x}+4e^{-3x}}$ and $I_2=\int \frac{dx}{e^{3x}+8e^x+4e^{-x}}$, then value of $I=I_1-2I_2$ is equal to

• A. $\frac{1}{2}ta{n}^{-1}\left(\frac{e^{2x}+2}{2e^x}\right)+c$
• B. $\frac{1}{2}ta{n}^{-1}\left(\frac{e^{2x}+1}{e^x-1}\right)+c$
• C. $\frac{1}{2}ta{n}^{-1}\left(\frac{e^x}{e^x+1}\right)+c$
• D. $\frac{1}{2}ta{n}^{-1}\left(\frac{e^x-1}{e^x+1}\right)+c$

Answer 1

The correct option is A $\frac{1}{2}ta{n}^{-1}\left(\frac{e^{2x}+2}{2e^x}\right)+c$

$I=I_1-2I_2$

$=\int \frac{dx}{e^x+\frac{8}{e^x}+\frac{4}{e^{3x}}}-2\int \frac{dx}{e^{3x}+8e^x+\frac{4}{e^x}}$

$=\int \frac{(e^{3x}-2e^x)}{e^{4x}+8e^{2x}+4}dxPut{e}^x=t\Rightarrow e^{x}dx=dt$

$I=\int \frac{(t^2-2)}{t^4+8t^2+4}dt$

$=\int \frac{(1-\frac{2}{t^2})}{{(t+\frac{2}{t})}^2+4}dt$

$t+\frac{2}{t}=z\Rightarrow (1-\frac{2}{t^2})dt=dz$

$I=\int \frac{dz}{x^2+4}=\frac{1}{2}ta{n}^{-1}\left(\frac{z}{2}\right)+c$

$=\frac{1}{2}ta{n}^{-1}\left(\frac{e^{2x}+2}{2e^z}\right)+c$