For x\geq0, \int\sqrt{x^2+2x} dx =

Correct Answer is : x+12√x2+2x−12Cosh−1(x+1)+C

{{x+1}/2}\sqrt{x^2+2x} + {1/2} \Sinh^{-1} {(x+1)/2}+\C

{{x+1}/2}\sqrt{x^2+2x} + {1/2} \Sinh^{-1} (x+1)+\C

{{x+1}/2}\sqrt{x^2+2x} - {1/2} \Cosh^{-1} {(x+1)/2}+\C

{{x+1}/2}\sqrt{x^2+2x} - {1/2} \Cosh^{-1} {(x+1)}+\C

TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Section: Mathematics

Question : 77 of 160

Marks: +1, -0

For x≥0, ∫√x2+2x dx =

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