evaluate ((lnx)^2)^ 1/2 dx

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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I suppose that you need to find the integral of the function `sqrt((lnx)^2) `  such that:

`int sqrt((lnx)^2) dx`

You need to remember that `sqrt(x^2) = |x|, `  hence `sqrt((lnx)^2) = |ln x|`

`` `int sqrt((lnx)^2) dx = int |ln x| dx`

You need to use integration by parts, hence you should write the formula such that:

`int udv = uv - int vdu`

You should consider `u = ln x =gt du = (dx)/x` , hence`dv = dx =gt v = x` Substituting `ln x`  for u, `(dx)/x`  for du, x for v and dx for dv yields:

int |ln x| dx = x*|ln x| - int x*(dx)/x

`int |ln x| dx = x*|ln x| - int dx`

`int |ln x| dx = x*|ln x| - x + c =gt int |ln x| dx = x*(|ln x| - 1) + ` c

`int |ln x| dx = x*(|ln x| - ln e) + c`

`int |ln x| dx = x*|ln(x/e)| + c`

Hence, evaluating the integral of function yields `int sqrt((lnx)^2) dx = x*|ln(x/e)| + c` .

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beckden | High School Teacher | (Level 1) Educator

Posted on

I guess you want to integrate this.

integral((ln(x))^2 1/2 dx)  = 1/2 integral(ln(x)*ln(x) dx)

integrate by parts

u = (ln(x))^2, du = 2 ln(x)/x dx  chain rule

dv = dx, v = x so

integral((ln(x)^2 1/2 dx) = 1/2 x (ln(x))^2 - 1/2 integral(x * 2 ln(x)/x dx)

1/2 integral(x * 2 ln(x)/x dx) = integral(ln(x) dx)

integral(ln(x) dx) integrate by parts

u = ln(x)   du = dx/x

dv = dx     v = x so

integral(ln(x) dx) = x ln(x) - integral(x dx/x) = x ln(x) - integral(dx) = x ln(x) - x

So combining

integral((ln(x)^2 1/2 dx) = 1/2 x (ln(x))^2 - 1/2 integral(x * 2 ln(x)/x dx)

= 1/2 x (ln(x))^2 - x ln(x) + x + C

We can check by differentiation.

d/dx (1/2 x ln(x))^2 - x ln(x) + x + C) =

= 1/2 (ln(x))^2 + 1/2 (x * 2 ln(x) / x) - ln(x) - x/x + 1

= 1/2 (ln(x))^2 + ln(x) - ln(x) - 1 + 1 = 1/2 (ln(x))^2

Which proves we have the correct answer.

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