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