KCET · Maths · Indefinite Integration
The value of \( \int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1} d x \) is equal to
- A \( e^{x} \tan ^{-1} x+c \)
- B \( \tan ^{-1}\left(e^{x}\right)+c \)
- C \( \tan ^{-1}\left(x^{e}\right)+c \)
- D \( e^{\tan ^{-1} x}+c \)
Answer & Solution
Correct Answer
(A) \( e^{x} \tan ^{-1} x+c \)
Step-by-step Solution
Detailed explanation
Given that, \( I=\int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1} d x \)
\( =\int e^{x}\left(\frac{\left(x^{2}+1\right) \tan ^{-1} x+1}{x^{2}+1}\right) d x \)
\( =\int e^{x}\left(\tan ^{-1} x+\frac{1}{x^{2}+1}\right) d x \)
Since \( \int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x) \)
So, \( I=e^{x} \tan ^{-1} x+c \)
\( =\int e^{x}\left(\frac{\left(x^{2}+1\right) \tan ^{-1} x+1}{x^{2}+1}\right) d x \)
\( =\int e^{x}\left(\tan ^{-1} x+\frac{1}{x^{2}+1}\right) d x \)
Since \( \int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x) \)
So, \( I=e^{x} \tan ^{-1} x+c \)
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