KCET · Maths · Indefinite Integration
\(\int \frac{(x-1) e^{x}}{(x+1)^{3}} d x\) is equal to
- A \(\frac{e^{x}}{x+1}+C\)
- B \(\frac{e^{x}}{(x+1)^{2}}+C\)
- C \(\frac{e^{x}}{(x+1)^{3}}+C\)
- D \(\frac{x \cdot e^{x}}{(x+1)}+C\)
Answer & Solution
Correct Answer
(B) \(\frac{e^{x}}{(x+1)^{2}}+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{x-1}{(x+1)^{3}} e^{x} d x\)
\(=\int \frac{(x+1)-2}{(x+1)^{3}} \cdot e^{x} d x\)
\(=\int \frac{e^{x}}{(x+1)^{2}} d x-2 \int \frac{e^{x}}{(x+1)^{3}} d x\)
\(=\frac{1}{(x+1)^{2}} \cdot e^{x}-\int \frac{-2}{(1+x)^{3}} \cdot e^{x} d x-2 \int \frac{e^{x}}{(x+1)^{3}} d x\)
\(=\frac{e^{x}}{(x+1)^{2}}+C\)
\(=\int \frac{(x+1)-2}{(x+1)^{3}} \cdot e^{x} d x\)
\(=\int \frac{e^{x}}{(x+1)^{2}} d x-2 \int \frac{e^{x}}{(x+1)^{3}} d x\)
\(=\frac{1}{(x+1)^{2}} \cdot e^{x}-\int \frac{-2}{(1+x)^{3}} \cdot e^{x} d x-2 \int \frac{e^{x}}{(x+1)^{3}} d x\)
\(=\frac{e^{x}}{(x+1)^{2}}+C\)
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