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Solve `int (ln^2(x))/x dx` ``
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High School Teacher
`int (ln^2 x)/x dx`
We will assume that y= lnx:
`lnx = y ==> ln^2 x = y^2 `
`==> dy = 1/x dx ==> dx = xdy`
Now we will substitute into the integral:
`==> int (ln^2 x)/dx dx = int y^2/x xdy = int y^2 dy= y^3/3 + C`
But we know that y= ln x
==> `int (ln^2 x)/x dx = (ln^3 x)/3 + C`
Posted by hala718 on October 26, 2011 at 9:57 PM (Answer #2)
`int (log^2x)/x dx` =`int log^2 d log x= (log^3x)/x+C`
Posted by oldnick on May 30, 2013 at 3:14 AM (Answer #3)
Substitute ln x by other variable. Instead of ln x put the variable y, in this way: ln x=y => ln^2(x)=y^2
You will see that if differentiate ln x with respect to x the result is 1/x.
d(ln x)/dx=dy => (1/x)*dx=dy
You will integrate:
Integral (ln^2(x))/x dx=Integral y^2 dy=y^3/3 + constant
Substitute y by ln x in result.
Integral (ln^2(x))/x dx= (ln^3(x))/3 + constant
Answer: Integral (ln^2(x))/x dx= (ln^3(x))/3 + constant
Posted by sciencesolve on October 26, 2011 at 6:59 PM (Answer #1)
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