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Question

If I=(a+xax+axa+x)dx, then I equals

A
2sin1(xa)+C
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B
2asin1(xa)+C
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C
2cos1(xa)+C
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D
2acos1(xa)+C
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Solution

The correct option is B 2asin1(xa)+C
I= a+xax+ axa+xdx

I=a+x+axa2x2dx

I=2aa2x2dx

We know that, 1a2x2dx=sin1(xa)+c

I=2asin1(xa)+c

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