CUET · MATHS · PYQ PAPER 2023
The value of the integral \(I=\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x\) is :
- A \(\frac{e^x}{1+x^2}+C\), where \(C\) is a constant
- B \(e^x \tan ^{-1} x+C\), where \(C\) is a constant
- C \(\frac{1}{1+x^2}+C\), where \(C\) is a constant
- D \(\tan ^{-1} x+C\), where \(C\) is a constant
Answer & Solution
Correct Answer
(B) \(e^x \tan ^{-1} x+C\), where \(C\) is a constant
Step-by-step Solution
Detailed explanation
\(I=\int e^x\left(f(x)+f'(x)\right) dx = e^x f(x) + C\) Let \(f(x) = \tan^{-1} x\) Then \(f'(x) = \frac{1}{1+x^2}\) \(I = e^x \tan^{-1} x + C\)
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