Question

\(\rm \int {4dx\over e^x + e^{-x}}\) = ?
1. 4 tan-1 ex+ C
2. 4 sin-1 ex + C
3. 4 cos-1 ex + C
4. 4 cot-1 ex + C

Answer 1

Correct Answer - Option 1 : 4 tan-1 ex+ C

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• \(\rm∫ {1\over 1+x^2} dx = \tan^{-1} x +C\)

 

Calculation:

I = \(\rm \int {4dx\over e^x + e^{-x}}\)

I = \(\rm 4\int {dx\over e^x + {1\over e^x}}\) 

I = \(\rm 4\int {e^x\over (e^x)^2 + 1}dx\)

Let ex = t ⇒ ex dx = dt

I = \(\rm 4\int {dt\over t^2+ 1}\)

I = 4 tan-1 t + C

Putting the value of t = ex

I = 4 tan-1 (ex) + C