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Question

Solve abxa2+x2dx

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Solution

Consider the given integral.


I=abxa2+x2dx …….. (1)



Let t=a2+x2


dtdx=0+2x


dt2=xdx



Therefore,


I=122a2a2+b21tdt


I=12[2t]2a2a2+b2


I=2a2a2+b2



Hence, this is the answer.


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