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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: problem no. 1
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little_genius (295)

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hello guys,


 from now onwards i ll be posting some very gud questions in maths  for all of us to try ......... so please all of u try and  enjoy.......


the headings of my problem post will be : "problem no. 1","problem no.2", etc/.......


problem no.1


find the sum of



and determine the value of x for which it is valid.........


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rudra.panda (2557)

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edited


God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~

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thedumbheadwithnobrain (887)

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log(1+((roota)x)2)=ax2-a2(x4/2)+a3(x6/3).......now change x to 1/x

log(1+((roota)/x)2)=a(1/x2)-a2(1/2x4)+a3(1/3x6).....



Add and get answer as log(1+((roota)/x)2)+log(1+((roota)x)2)


valid for x not equal to zero and valid for all x for a>0

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little_genius (295)

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ya the qures is rite

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