Correct Answer is : 2√ex−4−4tan−1(√ex−42)+C

\tan ^{-1}( \sqrt{e^{x}-4}}/{2})+\sqrt{e^{x}-4}+C

2 \sqrt{e^{x}-4}-4 \tan ^{-1}( \sqrt{e^{x}-4}}/{2})+C

2 \sqrt{e^{x}-4}-4 \cot ^{-1}( \sqrt{e^{x}-4}}/{2})+C

\sqrt{e^{x}-4}-4 \tan ^{-1}(\sqrt{e^{x}-4})+C

TS EAMCET 2015 Solved Paper

Section: Mathematics

Question : 8 of 160

Marks: +1, -0

∫√ex−4dx=

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