∫ x^2 to 1 (ln^2x/x)dx=

(A)(7/3e^2)
b)0
c)2.72
d)4.44
e)1.72

To solve the given integral ∫ x^2 to 1 (ln^2x/x) dx, we need to use integration by parts. Integration by parts is based on the product rule of differentiation.

Let's break down the steps to find the solution:

Step 1: Choose the proper parts
We need to choose two functions: u and dv, to apply the integration by parts formula. In this case, let's choose:

u = ln^2x (which can also be written as (lnx)^2)
dv = 1/x dx

Step 2: Compute the derivatives
To obtain the first derivative of u, we use the chain rule, which states:
d(u)/dx = (d(ln^2x)/dx)

Differentiating (ln^2x) with respect to x, we get:
d(u)/dx = 2lnx * (1/x)

To obtain the antiderivative of dv, we can simply integrate it:
v = ∫(dv) = ∫(1/x dx) = ln|x|

Step 3: Apply the integration by parts formula
The integration by parts formula is given by:
∫udv = uv - ∫vdu

Applying this formula, we have:
∫ x^2 to 1 (ln^2x/x) dx = (ln^2x * ln|x|) - ∫(2lnx * ln|x| / x) dx

Step 4: Simplify and evaluate the integral
Now our goal is to simplify the new integral:
∫(2lnx * ln|x| / x) dx

To simplify further, we can use a property of logarithms which states:
ln(a * b) = ln(a) + ln(b)

Using this property, we can rewrite the integral:
∫(2lnx * ln|x| / x) dx = 2∫(lnx * ln|x| / x) dx

Let's denote the new integral as I:
I = ∫(lnx * ln|x| / x) dx

To solve this integral, we can again use integration by parts.

Choose:
u = lnx
dv = ln|x| / x dx

Compute:
d(u)/dx = 1/x
v = ∫(ln|x| / x dx) = (1/2)(ln^2|x|)

Apply the integration by parts formula:
I = (lnx)(1/2)(ln^2|x|) - ∫((1/2)(ln^2|x|) / x) dx

Simplifying further:
I = (lnx)(1/2)(ln^2|x|) - (1/2)∫(ln^2|x| / x) dx

Notice that the second term in the last equation is the same integral we started with (I). So, substituting (I) in the equation, we get:
I = (lnx)(1/2)(ln^2|x|) - (1/2)I

Now, let's solve for I:
I + (1/2)I = (lnx)(1/2)(ln^2|x|)
(3/2)I = (lnx)(1/2)(ln^2|x|)
2I = 3(lnx)(ln^2|x|)

Finally, integrate I:
∫(lnx * ln|x| / x) dx = (3/2)(lnx)(ln^2|x|) + C

Substituting this result back into the original integral, we get:
∫ x^2 to 1 (ln^2x/x) dx = (ln^2x * ln|x|) - (3/2)(lnx)(ln^2|x|) + C

Now we can evaluate the definite integral by plugging in the limits of integration (x = 1 and x = x^2):

∫ x^2 to 1 (ln^2x/x) dx = [(ln^2(x^2) * ln|x^2|) - (3/2)(ln(x^2))(ln^2|x^2|)] - [(ln^2(1) * ln|1|) - (3/2)(ln(1))(ln^2|1|)]

Simplifying further:
∫ x^2 to 1 (ln^2x/x) dx = [(ln^2(x^2) * ln|x^2|) - (3/2)(ln(x^2))(ln^2|x^2|)] - 0

At this point, we can simplify the expressions inside the brackets and calculate the final result.