\int \sqrt{1+{\sqrt{x}}}/{x} \dx is equal to

Correct Answer is : 4√1+√x+2log(√1+√x−1√1+√x+1)+C

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

4 \sqrt{1+{\sqrt{x}}-2 \log ( \sqrt{1+{\sqrt{x}}-1}/\sqrt{1+{\sqrt{x}}+1}) +C

4 \sqrt{1+{\sqrt{x}}+2 \log ( \sqrt{1+{\sqrt{x}}-1}/\sqrt{1+{\sqrt{x}}+1}) +C

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

CUSAT 2022 BTech Solved Paper Shift 1

Question : 194 of 200

Marks: +1, -0

∫‌

√1+√x

‌dx is equal to

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