MHT CET · Maths · Indefinite Integration
The value of \(\int \mathrm{e}^x\left(\frac{x^2+4 x+4}{(x+4)^2}\right) \mathrm{d} x\) is
- A \(\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- B \(\mathrm{e}^x\left(\frac{4}{x+4}\right)+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- C \(\mathrm{e}^x\left(\frac{x}{(x+4)^2}\right)+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- D \(\mathrm{e}^x\left(\frac{4}{(x+4)^2}\right)+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Answer & Solution
Correct Answer
(A) \(\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int \mathrm{e}^x\left[\frac{x^2+4 x+4}{(x+4)^2}\right] \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{x(x+4)+4}{(x+4)^2}\right] \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{x}{x+4}+\frac{4}{(x+4)^2}\right] \mathrm{d} x\end{aligned}\)
\(=\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c} \)
\( \cdots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\)
\(=\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c} \)
\( \cdots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\)
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