KCET · Maths · Indefinite Integration
The value of \(\int \frac{x e^{x} d x}{(1+x)^{2}}\) is equal to
- A \(e^{x}(1+x)+C\)
- B \(e^{x}\left(1+x^{2}\right)+C\)
- C \(e^{x}(1+x)^{2}+C\)
- D \(\frac{e^{x}}{1+x}+C\)
Answer & Solution
Correct Answer
(D) \(\frac{e^{x}}{1+x}+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{x e^{x} d x}{(1+x)^{2}}\)
\(=\int \frac{e^{x}(x+1-1)}{(1+x)^{2}} d x\)
\(=\int e^{x}\left[\frac{1}{1+x}+\left(\frac{-1}{(1+x)^{2}}\right)\right] d x\)
Let \(\frac{1}{1+x}=f(x)\)
\(\therefore \quad f^{\prime}(x)=-\frac{1}{(1+x)^{2}}\)
Using the formula, \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\)
\(=e^{x} f(x)+C\)
\(\Rightarrow \quad I=e^{x}\left(\frac{1}{1+x}\right)+C=\frac{e^{x}}{1+x}+C\)
\(=\int \frac{e^{x}(x+1-1)}{(1+x)^{2}} d x\)
\(=\int e^{x}\left[\frac{1}{1+x}+\left(\frac{-1}{(1+x)^{2}}\right)\right] d x\)
Let \(\frac{1}{1+x}=f(x)\)
\(\therefore \quad f^{\prime}(x)=-\frac{1}{(1+x)^{2}}\)
Using the formula, \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\)
\(=e^{x} f(x)+C\)
\(\Rightarrow \quad I=e^{x}\left(\frac{1}{1+x}\right)+C=\frac{e^{x}}{1+x}+C\)
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