TS EAMCET · Maths · Indefinite Integration
If \(\int e^{-x} \tan ^{-1}\left(e^x\right) d x\) \(=f(x)-\frac{1}{2} \log \left(1+e^{2 x}\right)+C\), then \(f(x)\) equals
- A \(e^x-e^{-x} \tan ^{-1}\left(e^x\right)\)
- B \(x^2+e^{-x} \tan ^{-1}\left(e^x\right)\)
- C \(-e^{-x} \tan ^{-1}\left(e^x\right)\)
- D \(x-e^{-x} \tan ^{-1}\left(e^x\right)\)
Answer & Solution
Correct Answer
(D) \(x-e^{-x} \tan ^{-1}\left(e^x\right)\)
Step-by-step Solution
Detailed explanation
Let \(I=\int e^{-x} \tan ^{-1}\left(e^x\right) d x\) Put \(e^x=t\) \[ \begin{aligned} & \Rightarrow \quad e^x d x=d t \\ & \Rightarrow \quad d x=\frac{1}{e^x} d t \Rightarrow d x=\frac{1}{t} d t \end{aligned} \]…
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