CUET Preparation Today
$I=\int \frac{d x}{e^x+4 e^{-x}}=f(x)+c$ then f(x) is equal to |
$2 \tan ^{-1}\left(2 e^x\right)$ $\frac{1}{2} \tan ^{-1}\left(\frac{e^x}{2}\right)$ $2 \tan ^{-1} \frac{e^x}{2}$ $\frac{1}{2} \tan ^{-1}\left(2 e^{2 x}\right)$ |
$\frac{1}{2} \tan ^{-1}\left(\frac{e^x}{2}\right)$ |
$I=\int \frac{d x}{e^x+4 e^{-x}}=f(x)+c$ $\Rightarrow I=\int \frac{e^x d x}{e^{2 x}+4}$ Let $e^{x}=t \Rightarrow e^{x} dx=dt$ $\Rightarrow I=\int \frac{d t}{t^2+4}=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+c=\frac{1}{2} \tan ^{-1}\left(\frac{e^x}{2}\right)+c$ Hence (2) is the correct answer. |
