Q98 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
$\int_0^1 \dfrac{d x}{e^x+e^{-x}}=$
- Correct answerA$\tan ^{-1} e-\dfrac{\pi}{4}$
- B$\dfrac{\pi}{4}-\tan ^{-1} e$
- C$\tan e-\dfrac{\pi}{4}$
- D$\tan ^{-1} e$
Solution
Let $I = \int_{0}^{1} \dfrac{dx}{e^x + e^{-x}}$.
Multiply the numerator and denominator by $e^x$:
$I = \int_{0}^{1} \dfrac{e^x dx}{e^{2x} + 1}$.
Substitute $u = e^x$, so $du = e^x dx$. When $x=0$, $u=1$. When $x=1$, $u=e$.
$I = \int_{1}^{e} \dfrac{du}{u^2 + 1}$.
The integral of $\dfrac{1}{u^2 + 1}$ is $\tan^{-1}(u)$.
$I = [\tan^{-1}(u)]_{1}^{e} = \tan^{-1}(e) - \tan^{-1}(1)$.
Since $\tan^{-1}(1) = \dfrac{\pi}{4}$, we have $I = \tan^{-1}(e) - \dfrac{\pi}{4}$.
Answer: $\tan ^{-1} e-\dfrac{\pi}{4}$
Multiply the numerator and denominator by $e^x$:
$I = \int_{0}^{1} \dfrac{e^x dx}{e^{2x} + 1}$.
Substitute $u = e^x$, so $du = e^x dx$. When $x=0$, $u=1$. When $x=1$, $u=e$.
$I = \int_{1}^{e} \dfrac{du}{u^2 + 1}$.
The integral of $\dfrac{1}{u^2 + 1}$ is $\tan^{-1}(u)$.
$I = [\tan^{-1}(u)]_{1}^{e} = \tan^{-1}(e) - \tan^{-1}(1)$.
Since $\tan^{-1}(1) = \dfrac{\pi}{4}$, we have $I = \tan^{-1}(e) - \dfrac{\pi}{4}$.
Answer: $\tan ^{-1} e-\dfrac{\pi}{4}$
Hence, the correct answer is option: 'A'.