Question

intergrate (2x-(1/x^2))^2

can i intergrate without open up the square??

Answer 1

the normal way is opening up the square and intergrate. can i intergrate without open up the square??

Answer 2

\[\int\limits_{}^{}(2x-(1/x^2))^2\]

Answer 3

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

Answer 4

???

Answer 5

inequalities?? I'm not ask abt that.. I'm asking abt intergration

Answer 6

why bother? it is easiest to integrate
\[\int 4x^4-\frac{4}{x}+\frac{1}{x^4}dx\]

Answer 7

ok thx

Answer 8

no its
\[\int\limits_{}^{}4x^2-4/x+1/x^4dx\]

Answer 9

\[(2x−(1/x^2))^2=4x^2-4/x+1/x^4\]