Q60 / 150 · MHT CET 2025 (20 Apr Shift 2) · Mathematics

$\int \frac{e^{\tan ^{-1} 2 x}}{1+4 x^2}=$

  • A
    $4 \mathrm{e}^{\tan ^{-1} 2 x}+\mathrm{c}, \quad$ where c is the constant of integration
  • B
    $e^{\tan ^{-1} 2 x}+c, \quad$ where $c$ is the constant of integration
  • Correct answer
    C
    $\frac{e^{\tan ^{-1} 2 x}}{2}+c, \quad$ where $c$ is the constant of integration
  • D
    $2 \mathrm{e}^{\tan ^{-1} 2 x}+\mathrm{c}, \quad$ where c is the constant of integration

Solution

The integral $\int \frac{e^{\tan^{-1} 2x}}{1+4x^2} dx$ is evaluated using substitution.

Let $u = \tan^{-1} 2x$. The derivative is $\frac{du}{dx} = \frac{2}{1+4x^2}$, so $\frac{1}{1+4x^2} dx = \frac{1}{2} du$.

Substituting gives $\int e^u \cdot \frac{1}{2} du = \frac{1}{2} \int e^u du = \frac{1}{2} e^u + C$.

Replacing $u$ yields the final result: $\frac{1}{2} e^{\tan^{-1} 2x} + C$.

This corresponds to option C.

Hence, the correct answer is option: 'C'.

Relevant Questions