Q64 / 150 · MHT CET 2025 (25 Apr Shift 1) · Mathematics

$\int \mathrm{e}^x\left(\frac{x+5}{(x+6)^2}\right) \mathrm{d} x$ is

  • A
    $\frac{\mathrm{e}^x}{(x+6)^2}+\mathrm{c}$, where c is the constant of integration.
  • B
    $\frac{e^x}{x+5}+c,$ where $c$ is the constant of integration.
  • C
    $\frac{\mathrm{e}^x}{(x+5)^2}+\mathrm{c}$, where c is the constant of integration.
  • Correct answer
    D
    $\frac{\mathrm{e}^x}{x+6}+\mathrm{c},$ where c is the constant of integration.

Solution

Evaluation of the integral $\int e^x \left( \frac{x+5}{(x+6)^2} \right) \, dx$ relies on recognizing it as $\int e^x (f(x) + f'(x)) \, dx$, which integrates to $e^x f(x) + c$.

Express the integrand as $\frac{x+5}{(x+6)^2} = \frac{(x+6)-1}{(x+6)^2} = \frac{1}{x+6} - \frac{1}{(x+6)^2}$.

Let $f(x) = \frac{1}{x+6}$, so $f'(x) = -\frac{1}{(x+6)^2}$. The integrand matches $e^x (f(x) + f'(x))$, yielding the integral.

Thus, $\int e^x \left( \frac{1}{x+6} - \frac{1}{(x+6)^2} \right) \, dx = e^x \cdot \frac{1}{x+6} + c$.

$e^x \left( \frac{1}{x+6} \right) + c$ corresponds to option D.

Final answer: $\boxed{\text{D}}$

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

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