RE: Math Problem 03-31-2013, 07:50 PM
#4
You obviously mean 1/(6x^2), which is NOT (1/6)x^2
1/(6x^2)
= 1/6 * (x^-2)
Now the integral of that:
1/6 * (-1) x^(-2 + 1) + c
=-(1/6) * (x^-1) + c
=-1/(6*x) + c
1/(6x^2)
= 1/6 * (x^-2)
Now the integral of that:
1/6 * (-1) x^(-2 + 1) + c
=-(1/6) * (x^-1) + c
=-1/(6*x) + c
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