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use integration by parts to integrate integration sign, ln x / square root x, whole...
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then by differenciation;
dx = 2tdt
`int(lnx/sqrtx)dx = int(lnt^2/sqrtt^2)*2tdt`
= `4intlnt dt`
Posted by jeew-m on July 26, 2012 at 1:57 AM (Answer #1)
High School Teacher
int ln(x)/sqrt(x) dx
use LIPET to find to use `u=ln(x)` , `dv = 1/sqrt(x) dx`
This gives `du = 1/x dx` and `v = 2sqrt(x)`
Now apply integration by parts
`int udv = uv - int v du` to get
`int ln(x)/sqrt(x) dx = ln(x) (2sqrt(x)) - int 2sqrt(x) 1/x dx`
`int ln(x)/sqrt(x) dx = 2sqrt(x)ln(x) - 2 int sqrt(x)/x dx`
Since `sqrt(x)/x = x^(1/2)/x^1 = x^(-1/2)`
`int ln(x)/sqrt(x) dx = 2sqrt(x)ln(x) - 2 int x^(-1/2) dx`
And finally we get
`int ln(x)/sqrt(x) dx = 2sqrt(x)ln(x) - 4sqrt(x) + C` which is the same as the answer above but without using a table of integrals.
Posted by beckden on July 26, 2012 at 2:55 AM (Answer #2)
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