MHT CET · Maths · Indefinite Integration
\(\int \mathrm{e}^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} \mathrm{~d} x=\)
- A \(\mathrm{e}^{x}\left(\frac{1}{1+x^{2}}\right)+\mathrm{C}\)
- B \(\mathrm{e}^{x}\left(\frac{-1}{1+x^{2}}\right)+\mathrm{C}\)
- C \(\mathrm{e}^{x}\left(\frac{2}{1+x^{2}}\right)+\mathrm{C}\)
- D \(\mathrm{e}^{x}\left(\frac{-2}{1+x^{2}}\right)+\mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{e}^{x}\left(\frac{1}{1+x^{2}}\right)+\mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(I=\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x\)
\(=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x=\int e^{x}[\frac{1+x^{2}}{\left(1+x^{2}\right)^{2}}\) \(-\frac{2 x}{\left(1+x^{2}\right)^{2}}] d x\)
\(=\int e^{x}\left[\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x=\frac{e^{x}}{1+x^{2}}+C\)
\(=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x=\int e^{x}[\frac{1+x^{2}}{\left(1+x^{2}\right)^{2}}\) \(-\frac{2 x}{\left(1+x^{2}\right)^{2}}] d x\)
\(=\int e^{x}\left[\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x=\frac{e^{x}}{1+x^{2}}+C\)
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