CUET · MATHS · PYQ PAPER 2023
\(\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x\) is equal to :
- A \(e^x\left(\frac{1}{1+x^2}\right)+C\)
- B \(\tan ^{-1} x+C\)
- C \(e^x \tan ^{-1} x+C\)
- D \(e^x \cot ^{-1} x+C\)
Answer & Solution
Correct Answer
(C) \(e^x \tan ^{-1} x+C\)
Step-by-step Solution
Detailed explanation
\(\int e^x (f(x)+f'(x)) dx = e^x f(x) + C\) \(\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x = e^x \tan ^{-1} x+C\)
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