First slide

Methods of integration

Question

$\int \frac{e^{2 x} - e^{- 2 x}}{e^{2 x} + e^{- 2 x}} dx$ is equal to

Easy

A

$\log |e^{2 x} - e^{- 2 x}| + C$
B

$\frac{1}{2} \log |e^{2 x} + e^{- 2 x}| + C$
C

$\log |e^{2 x} + e^{- 2 x}| + C$
D

None of these

Solution

$\begin{array}{l} \int \frac{e^{2 x} - e^{- 2 x}}{e^{2 x} + e^{- 2 x}} dx \\ Let e^{2 x} + e^{- 2 x} = t \\ \Rightarrow 2 e^{2 x} - 2 e^{- 2 x} = \frac{dt}{dx} \\ \Rightarrow dx = \frac{dt}{2 (e^{2 x} - e^{- 2 x})} \\ \begin{aligned} e^{2 x} - e^{- 2 x} & e^{2 x} + e^{- 2 x} dx = \int \frac{e^{2 x} - e^{- 2 x}}{t} \frac{dt}{2 [e^{2 x} - e^{- 2 x}]} \end{aligned} \\ = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log | t | + C \\ = \frac{1}{2} \log |e^{2 x} + e^{- 2 x}| + C \end{array}$

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