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Question

If x19dxx5(x5x10x10)=x30m+(x201)3/2n+C, where C is arbitrary constant of integration and m,nN, then the value of (m+n) is

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Solution

I=x19dxx5(x5x10x10)
=x19dxx10x201
=x19(x10+x201)dx
=(x29+x19x201)dx
=x3030+(x19x201)dx
Put x201=t220x19dx=2tdt
I=x3030+110t2dt
=x3030+130t3+C
=x3030+(x201)3/230+C
m+n=30+30=60

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