∫ 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 integral ∫x^2 to 1 (ln^2x/x)dx, we can use integration by parts. The formula for integration by parts is:
∫u dv = uv - ∫v du
Let's choose u = ln^2x and dv = 1/x dx. Then, we can find du and v as follows:
Differentiating u = ln^2x
du = (2lnx) * (1/x) dx
du = 2lnxdx/x
Integrating dv = 1/x dx
v = ln(x)
Using the formula for integration by parts, we can rewrite the integral as:
∫x^2 to 1 (ln^2x/x)dx = uv - ∫v du
Plugging in the values, we have:
∫x^2 to 1 (ln^2x/x)dx = ln^2x * ln(x) - ∫ln(x) * 2lnxdx/x
Simplifying the integral, we have:
∫x^2 to 1 (ln^2x/x)dx = ln^2x * ln(x) - 2∫ln^2xdx
To find the integral of ln^2x, we can use integration by parts again. Let's choose u = ln^2x and dv = dx. Then, we can find du and v as follows:
Differentiating u = ln^2x
du = (2lnx) * 1/x dx
du = 2lnxdx/x
Integrating dv = dx
v = x
Using the formula for integration by parts, we can rewrite the integral as:
∫ln^2xdx = uv - ∫v du
∫ln^2xdx = ln^2x * x - ∫x * (2lnxdx/x)
Simplifying the integral, we have:
∫ln^2xdx = ln^2x * x - 2∫lndx
The integral of ln(x) can be found by integrating by parts one more time. Let's choose u = ln(x) and dv = dx. Then, we can find du and v as follows:
Differentiating u = ln(x):
du = 1/x dx
Integrating dv = dx:
v = x
Using the formula for integration by parts, we have:
∫lndx = uv - ∫v du
∫lndx = ln(x) * x - ∫x * 1/x dx
∫lndx = ln(x) * x - ∫dx
∫lndx = ln(x) * x - x
Now we can substitute this result back into the previous equation:
∫ln^2xdx = ln^2x * x - 2 * (ln(x) * x - x)
∫ln^2xdx = ln^2x * x - 2ln(x) * x + 2x
Finally, we evaluate the original integral:
∫x^2 to 1 (ln^2x/x)dx = ln^2(1) * 1 - 2ln(1) * 1 + 2(1) - [ln^2(x) * x - 2ln(x) * x + 2x]
∫x^2 to 1 (ln^2x/x)dx = 0 - 0 + 2 - [ln^2(1) * 1 - 2ln(1) * 1 + 2(1)]
∫x^2 to 1 (ln^2x/x)dx = 2 - [0 - 0 + 2]
∫x^2 to 1 (ln^2x/x)dx = 2 - 2
∫x^2 to 1 (ln^2x/x)dx = 0
Therefore, the answer is b) 0.