TS EAMCET · Maths · Indefinite Integration
\(\int \sqrt{e^x-4} d x\) equals
- A \(\tan ^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right)+\sqrt{e^x-4}+C\)
- B \(2 \sqrt{e^x-4}-4 \tan ^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right)+C\)
- C \(2 \sqrt{e^x-4}-4 \cot ^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right)+C\)
- D \(\sqrt{e^x-4}-4 \tan ^{-1}\left(\sqrt{e^x-4}\right)+C\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{e^x-4}-4 \tan ^{-1}\left(\frac{\sqrt{e^x-4}}{2}\right)+C\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Let } I=\int \sqrt{e^x-4} d x \\ & \begin{aligned} \text { Put } e^x-4 & =t^2 \\ \Rightarrow \quad e^x d x & =2 t d t \\ \Rightarrow \quad d x & =\frac{2 t}{t^2+4} d t \\ \therefore \quad I & =\int t \cdot \frac{2 t}{t^2+4} d t \\ & =2 \int…
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