Question

\[ \int \frac{\mathrm{e}^{\tan ^{-1} 2 x}}{1+4 x^2}= \]

• A. \(4\mathrm{e}^{\mathrm{tan}^{-1}2x}+\mathrm{c}\), where c is the constant of integration
• B. \(\mathrm{e}^{\mathrm{tan}^{-1}2x}+\mathrm{c}\), where c is the constant of integration
• C. \(\frac{\mathrm{e}^{\mathrm{tan}^{-1}2x}}{2}+\mathrm{c}\), where c is the constant of integration
• D. \(2\mathrm{e}^{\mathrm{tan}^{-1}2x}+\mathrm{c}\), where c is the constant of integration

Hint:

Substitute \(t = \tan^{-1}2x\), so \(dt = \dfrac{2}{1+4x^2}dx\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The integrand has \(e^{\tan^{-1}2x}\) and the derivative of \(\tan^{-1}2x\) is \(\dfrac{2}{1+4x^2}\), so substitution works.

Step 2: Substitute
Let \(t = \tan^{-1}2x\). Then \(dt = \dfrac{2}{1+4x^2}\,dx\), so \(\dfrac{dx}{1+4x^2} = \dfrac{dt}{2}\).

Step 3: Integrate
\(\displaystyle\int e^{t}\,\dfrac{dt}{2} = \dfrac{e^{t}}{2} + c\)

Step 4: Back substitute
The result is \(\dfrac{e^{\tan^{-1}2x}}{2} + c\).

Final Answer:
The integral equals \(\dfrac{e^{\tan^{-1}2x}}{2} + c\). \[ \boxed{\text{(C) }\dfrac{e^{\tan^{-1}2x}}{2}+c} \]