$\int \frac{4 e^{x}}{2 e^{x} - 5 e^{- x}} d x$ is equal to

4 log | e^{x} - 5 | + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{4} log ∣ ∣ e^{2 x} - 5 ∣ ∣ + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

log | 2 e^{x} - 5 e^{- x} | + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 log | 2 e^{x} - 5 | + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

log ∣ ∣ 2 e^{2 x} - 5 ∣ ∣ + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $log | 2 e^{x} - 5 e^{- x} | + C$
Consider, $I = \int \frac{4 e^{x}}{2 e^{x} - 5 e^{- x}} d x$

$= \int \frac{4 e^{x}}{2 e^{x} \cdot - \frac{5}{e^{x}}} d x$

$= \int \frac{4 e^{2 x}}{2 e^{2 x} - 5} d x$

Put $2 e^{2 x} - 5 = t$

$\Rightarrow 4 e^{2 x} d x = d t$

$∴ I = \int \frac{d t}{t}$

$= log | t | + C$

$\Rightarrow I = log ∣ ∣ 2 e^{2 x} - 5 ∣ ∣ + C$

Suggest Corrections

Similar questions

\int (\frac{4 e^{x} - 25}{2 e^{x} - 5}) d x = A x + B log | 2 e^{x} - 5 | + c

\int \frac{e^{x}}{e^{2 x} + 6 e^{x} + 5} d x .

= \frac{1}{4} l o g ∣ ∣ ∣ \frac{e^{x} + 1}{e^{x} + k} ∣ ∣ ∣ + C

\int \frac{2 e^{x} + 3 e^{- x}}{4 e^{x} + 7 e^{- x}} d x = \frac{1}{14} (u x + v {log}_{e} (4 e^{x} + 7 e^{- x})) + C

C

u + v

\int (e^{2 x} + 2 e^{x} - e^{- x} - 1) e^{(e^{x} + e^{- x})} d x = g (x) e^{(e^{x} + e^{- x})} + c

c

g (0)

\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = a x + b ln (4 e^{x} + 5 e^{- x}) + C

Join BYJU'S Learning Program