anonymous
  • anonymous
intergrate (2x-(1/x^2))^2 can i intergrate without open up the square??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
the normal way is opening up the square and intergrate. can i intergrate without open up the square??
anonymous
  • anonymous
\[\int\limits_{}^{}(2x-(1/x^2))^2\]
anonymous
  • anonymous
here's a hint!! The expression a < b is read as a is less than b while the expression a > b is read as a is greater than b. The < and > signs define what is known as the sense of the inequality (indicated by the direction of the sign). Two inequalities are said to have (a) the same sense if the signs of inequality point in the same direction; and (b) the opposite sense if the signs of inequality point in the opposite direction. here is a example The inequalities x + 3 > 2 and x + 1 > 0 have the same sense. So do the inequalities 3x - 1 < 4 and x2- 1 < 3

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anonymous
  • anonymous
???
anonymous
  • anonymous
inequalities?? I'm not ask abt that.. I'm asking abt intergration
anonymous
  • anonymous
why bother? it is easiest to integrate \[\int 4x^4-\frac{4}{x}+\frac{1}{x^4}dx\]
anonymous
  • anonymous
ok thx
anonymous
  • anonymous
no its \[\int\limits_{}^{}4x^2-4/x+1/x^4dx\]
anonymous
  • anonymous
\[(2x−(1/x^2))^2=4x^2-4/x+1/x^4\]

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