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Question

Prove that (a2x2)dx=x(a2x2)2+a22sin1xa

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Solution

I=a2x2dx
Integration by parts differentiate a2x2 and integrate 1
I=a2x2dxda2x2dx(dx)dx
I=xa2x2122xa2x2x;dx
I=xa2x2+x2a2+a2a2x2dx
I=xa2x2a2x2a2x2dx+a2a2x2dx
I=xa2x2a2x2dx+a2dxa2x2dx
I=xa2x2I+a2sin1xa
2I=xa2x2+a2sin1(x2)
I=x2a2x2+a22sin1(xa)
Hence, solved.


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