KCET · Maths · Definite Integration
\(\int_0^1 \frac{x e^x}{(2+x)^3} d x\) is equal to
- A \(\frac{1}{27} \cdot e-\frac{1}{8}\)
- B \(\frac{1}{27} \cdot e+\frac{1}{8}\)
- C \(\frac{1}{9} \cdot e+\frac{1}{4}\)
- D \(\frac{1}{9} \cdot e-\frac{1}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{9} \cdot e-\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Let
Here, \(\frac{1}{(2+x)^2}=f(x)\) and \(\frac{-2}{(2+x)^3}=f^{\prime}(x)\)
We know, \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c\)
\[
l=\left[e^x \cdot \frac{1}{(2+x)^2}\right]_0^1=\frac{e^x}{(2+1)^2}-\frac{e^0}{(2+0)^2}=\frac{e}{9}-\frac{1}{4}
\]
Here, \(\frac{1}{(2+x)^2}=f(x)\) and \(\frac{-2}{(2+x)^3}=f^{\prime}(x)\)
We know, \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c\)
\[
l=\left[e^x \cdot \frac{1}{(2+x)^2}\right]_0^1=\frac{e^x}{(2+1)^2}-\frac{e^0}{(2+0)^2}=\frac{e}{9}-\frac{1}{4}
\]
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