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Q.

For any natural number m,

x7m+x2m+xm2x6m+7xm+141/mdx, where x>0 equals

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a

7x7m+2x2m+14xmnm+1m14(m+1)+C

b

7x7m+2x2m+xmm+1m14(m+1)+C

c

2x7m+7x2m+14xmm+1m14(m+1)+C

d

2x7m+14x2m+7xmnm+114(m+1)+C

answer is C.

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Detailed Solution

Let I=x7m+x2m+xm2x6m+7xm+141/mdx

Then,

I=x7m1+x2m1+xm12x7m+7x2m+14xm1/mdx

Let 2x7m+7x2m+14xm=t. Then,

14mx7m1+x2m1+xm1dx=dt l=114mt1/mdt=114mt1m+11m+1+C I=114(m+1)2x7m+7x2m+14xmm+1m+C

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